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Solving 2nd Degree Equations ax2 + bx + c = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
Solving 2nd Degree Equations ax2 + bx + c = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = 7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d.
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3
7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
exact answers
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0
The Square-Root Method
Use the square-root method if there is no x-term.
I. Solve for the x2 and get the form x2 = d
II. Then x = Β±οƒ–d are the roots.
Example A. Solve 3x2 – 7 = 0
3x2 – 7 = 0 solve for x2
3x2 = 7
x2 = take square root
x = Β±οƒ–7/3 ο‚» Β±1.53
7
3
exact answers
approx. answers
To solve ax2 + bx + c = 0:
1st: Use the square-root method if the x-term is missing.
2nd: Try factoring it into two binomials.
3rd: Use the quadratic formula (QF).
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Solving 2nd Degree Equations ax2 + bx + c = 0
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
3x2 – 2x – 8 = 0
(3x + 4)(x – 2) = 0
Example B. Which of the following equations is factorable?
Solve it by factoring if it is.
a. 3x2 – 2x + 8 = 0
a = 3, b = –2, c =8
so b2 – 4ac
= (–2)2 – 4(3)(8)
= 4 – 96 = –92
Not a perfect square
so it's not factorable
Factoring Method
Factor the equation into the form (#x + #)(#x + #) = 0.
Hint: Find b2 – 4ac, if it is not a perfect square then the
equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…)
Solving 2nd Degree Equations ax2 + bx + c = 0
b. 3x2 – 2x – 8 = 0
a = 3, b = –2, c = –8
b2 – 4ac = (–2)2 – 4(3)(-8)
= 4 + 96 = 100 = 102
So its factorable.
3x2 – 2x – 8 = 0
(3x + 4)(x – 2) = 0
x = -4/3, x = 2
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28.
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
=
1 Β± οƒ–7
3
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac
2a
Solving 2nd Degree Equations ax2 + bx + c = 0
Example C. Solve 3x2 – 2x – 2 = 0
Check if it is factorable: a = 3, b = –2, c = –2
So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable!
Use QF, we get
x = – (–2) Β± οƒ– 28
2(3)
=
2 Β± οƒ– 28
6
=
2 Β± 2οƒ–7
6
=
1 Β± οƒ–7
3
ο‚» {1.22
–0.549
Use the quadratic formula (QF)
The roots for the equation ax2 + bx + c = 0 are
x =
b2 – 4ac is called the discriminant because its value
indicates what type of roots there are. Specifically,
if b2 – 4ac is a perfect square, we have fractional roots,
if b2 – 4ac < 0 there is no real roots.
Example D. Solve 3x2 – 2x + 2 = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
–20 is not a real number, so there is no real solution.
Solving 2nd Degree Equations ax2 + bx + c = 0
Example D. Solve 3x2 – 2x + 2 = 0
To check if it is factorable; a = 3, b = -2, c = -2
So b2 – 4ac = (–2)2 – 4(3)(2) = –20,
–20 is not a real number, so there is no real solution.
Solving 2nd Degree Equations ax2 + bx + c = 0
Solving 2nd Degree Equations ax2 + bx + c = 0
1. 3x2 – 12 = 0
Exercise A. Solve. Use the square root methods.
Give both the exact answers and the numerical answers.
If the answer is not real, state so.
2. 3x2 – 15 = 0 3. 3x2 + 15 = 0
4. x2 – 3 = –x2 + 15 5. –6 = 4x2 – 15 6. (x – 2)2 = 2
7. 4 = (2x – 3)2
Exercise B. Solve by factoring and by the quadratic formula.
14. x2 – 3x = 10 15. x(x – 2) = 24 16. 2x2 = 3(x + 1) – 1
8. x2 – 3x = 4 9. x2 – 15 = 2x 10. x2 + 5x + 12 = 0
11. –x2 – 2x + 8 = 0 12. 9 – 3x – 2x2 = 0 13. 2x2 – x – 1 = 0
Exercise C. Solve by the quadratic formula.
17. x2 – x + 1 = 0 18. x2 – x – 1 = 0
19. x2 – 3x – 2 = 0 20. x2 – 2x + 3 = 0
If the answers are not real numbers, state so.
21. 2x2 – 3x – 1 = 0 22. 3x2 = 2x + 3
Solving 2nd Degree Equations ax2 + bx + c = 0
Exercise C. Solve by the quadratic formula.
If the answers are not real numbers, state so.
23. 2(x2 – 1) + x = 4 24. (x – 1)(x + 1) = 2x(x + 2)
25.
(x – 1)
(x + 2)
=
2x
(x + 2)
26.
(x + 1)
(x + 2)
=
(x + 2)
(2x + 1)
1
Π€ 1 – Π€
1 – Π€
Π€1
Π€ =
27. Cut a stick of unit length (1) into two parts in a manner
such that β€œthe ratio of the whole to the large part is the same
as the ratio of the large part to the small part”
In picture,
the golden ratio Π€ (phi) is the ratio 1 : Π€. Find the exact and
approximate value of Π€. (Google β€œgolden ratio” )
Solving 2nd Degree Equations ax2 + bx + c = 0
HW answers
1. x = Β±2 3. no real answers 5. x = Β±3/2
7. x = Β½, 5/2 9. x = –3, 5 11. x = –4, 2
17. no real answer 19. x =
13. x = –1/2, 1 15. x = 6, –4
3 ±√17
2 21. x =
3 ±√17
4
23. x = –2, 3
27. Π€ = –1 +√5
2
25. x = –1, –2
approx. 0.618

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1.1 review solving 2nd degree equations

  • 1. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 2. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
  • 3. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.
  • 4. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 5. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 6. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 7. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 8. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 9. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 10. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 11. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 12. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 13. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 14. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 15. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers approx. answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 16. Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 17. Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0
  • 18. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 19. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 20. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 21. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 22. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0
  • 23. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8
  • 24. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8)
  • 25. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100
  • 26. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102
  • 27. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable.
  • 28. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable. 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0
  • 29. Example B. Which of the following equations is factorable? Solve it by factoring if it is. a. 3x2 – 2x + 8 = 0 a = 3, b = –2, c =8 so b2 – 4ac = (–2)2 – 4(3)(8) = 4 – 96 = –92 Not a perfect square so it's not factorable Factoring Method Factor the equation into the form (#x + #)(#x + #) = 0. Hint: Find b2 – 4ac, if it is not a perfect square then the equation is not factorable. (perfect squares: 0, 1, 4, 9, 16,…) Solving 2nd Degree Equations ax2 + bx + c = 0 b. 3x2 – 2x – 8 = 0 a = 3, b = –2, c = –8 b2 – 4ac = (–2)2 – 4(3)(-8) = 4 + 96 = 100 = 102 So its factorable. 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = -4/3, x = 2
  • 30. Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF)
  • 31. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x =
  • 32. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are.
  • 33. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots,
  • 34. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 35. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 36. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 37. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 38. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 39. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 40. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 41. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 42. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 43. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 ο‚» {1.22 –0.549 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 44. Example D. Solve 3x2 – 2x + 2 = 0 Solving 2nd Degree Equations ax2 + bx + c = 0
  • 45. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 Solving 2nd Degree Equations ax2 + bx + c = 0
  • 46. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, Solving 2nd Degree Equations ax2 + bx + c = 0
  • 47. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, –20 is not a real number, so there is no real solution. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 48. Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable; a = 3, b = -2, c = -2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20, –20 is not a real number, so there is no real solution. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 49. Solving 2nd Degree Equations ax2 + bx + c = 0 1. 3x2 – 12 = 0 Exercise A. Solve. Use the square root methods. Give both the exact answers and the numerical answers. If the answer is not real, state so. 2. 3x2 – 15 = 0 3. 3x2 + 15 = 0 4. x2 – 3 = –x2 + 15 5. –6 = 4x2 – 15 6. (x – 2)2 = 2 7. 4 = (2x – 3)2 Exercise B. Solve by factoring and by the quadratic formula. 14. x2 – 3x = 10 15. x(x – 2) = 24 16. 2x2 = 3(x + 1) – 1 8. x2 – 3x = 4 9. x2 – 15 = 2x 10. x2 + 5x + 12 = 0 11. –x2 – 2x + 8 = 0 12. 9 – 3x – 2x2 = 0 13. 2x2 – x – 1 = 0 Exercise C. Solve by the quadratic formula. 17. x2 – x + 1 = 0 18. x2 – x – 1 = 0 19. x2 – 3x – 2 = 0 20. x2 – 2x + 3 = 0 If the answers are not real numbers, state so. 21. 2x2 – 3x – 1 = 0 22. 3x2 = 2x + 3
  • 50. Solving 2nd Degree Equations ax2 + bx + c = 0 Exercise C. Solve by the quadratic formula. If the answers are not real numbers, state so. 23. 2(x2 – 1) + x = 4 24. (x – 1)(x + 1) = 2x(x + 2) 25. (x – 1) (x + 2) = 2x (x + 2) 26. (x + 1) (x + 2) = (x + 2) (2x + 1) 1 Π€ 1 – Π€ 1 – Π€ Π€1 Π€ = 27. Cut a stick of unit length (1) into two parts in a manner such that β€œthe ratio of the whole to the large part is the same as the ratio of the large part to the small part” In picture, the golden ratio Π€ (phi) is the ratio 1 : Π€. Find the exact and approximate value of Π€. (Google β€œgolden ratio” )
  • 51. Solving 2nd Degree Equations ax2 + bx + c = 0 HW answers 1. x = Β±2 3. no real answers 5. x = Β±3/2 7. x = Β½, 5/2 9. x = –3, 5 11. x = –4, 2 17. no real answer 19. x = 13. x = –1/2, 1 15. x = 6, –4 3 ±√17 2 21. x = 3 ±√17 4 23. x = –2, 3 27. Π€ = –1 +√5 2 25. x = –1, –2 approx. 0.618