# DETERMINE IF ORDERED PAIRS ARE SOLUTIONS TO A QUADRATIC FUNCTIONS

To check if the given ordered pair is a solution, we have to follow the given steps.

Step 1 :

Consider the given solution ordered pair as (x, y)

Step 2 :

Apply the values of x and y, in the given quadratic function.

Step 3 :

• Incase the given ordered pair is satisfying the given quadratic function, the ordered pair is a solution.
• Incase the given ordered pair is not satisfying the given quadratic function, the ordered pair is not a solution.

State whether the following quadratic functions are satisfied by the given ordered pairs

Problem 1 :

y = 3x2 + 2x        (2, 16)

Solution :

y = 3x2 + 2x

Given ordered pair is (2, 16)

16 = 3(2)2 + 2(2)

16 = 3(4) + 4

16 = 12 + 4

16 = 16

Since the given ordered pair is satisfying the given quadratic function, it is a solution.

Problem 2:

f(x) = -x2 - 2x + 1        (-3, 1)

Solution :

f(x) = -x2 - 2x + 1

Given ordered pair is (-3, 1)

f(-3) = -(-3)2 - 2(-3) + 1

1 = -9 + 6 + 1

1 = -9 + 7

≠ -2

Since the given ordered pair is not satisfying the given quadratic function, it is not a solution.

Problem 3 :

f(x) = 5x2 - 10        (0, 5)

Solution :

f(x) = 5x2 - 10

Given ordered pair is (0, 5)

f(0) = 5(0)2 - 10

5 = 0 - 10

≠ -10

So, the given ordered pair (0, 5) is not a solution.

Problem 4 :

f(x) = 2x2 + 5x - 3        (4, 9)

Solution :

f(x) = 2x2 + 5x - 3

Given ordered pair is (4, 9)

f(4) = 2(4)2 + 5(4) - 3

9 = 2(16) + 20 - 3

9 = 32 + 20 - 3

9 = 52 - 3

9 49

So, the given ordered pair (4, 9) is not a solution.

Problem 5 :

f(x) = -2x2 + 3x        (-1/2, 1)

Solution :

f(x) = -2x2 + 3x

Given ordered pair is (-1/2, 1)

f(-1/2) = -2(-1/2)2 + 3(-1/2)

1 = -2(1/4) - (3/2)

1 = -1/2 - (3/2)

1 = (-1-3)/2

1 = -4/2

1 -2

So, the given ordered pair (-1/2, 1) is not a solution.

Problem 6 :

f(x) = -7x2 + 8x + 15        (-1, 16)

Solution :

f(x) = -7x2 + 8x + 15

Given ordered pair is (-1, 16)

f(-1) = -7(-1)2 + 8(-1) + 15

16 = -7(1) - 8 + 15

16 = -7 - 8 + 15

16 = -15 + 15

16 0

So, (-1, 16) is not a solution.

Problem 7 :

f(x) = 3x2 - 13x + 4        (2, -10)

Solution :

f(x) = 3x2 - 13x + 4

Given ordered pair is (2, -10)

f(2) = 3(2)2 - 13(2) + 4

-10 = 3(4) - 26 + 4

-10 = 12 - 26 + 4

-10 = 16 - 26

-10 = -10

(2, -10) is a solution.

Find the values of x for the given value of y for each of the following quadratic function.

Problem 8 :

y = x2 + 6x + 10     {y = 1}

Solution :

y = x2 + 6x + 10

Applying y = 1, we get

1 = x2 + 6x + 10

x2 + 6x + 10 - 1 = 0

x2 + 6x + 9 = 0

x2 + 3x + 3x + 9 = 0

x(x + 3) + 3(x + 3) = 0

(x + 3)(x + 3) = 0

x = -3 and x = -3

The solutions are (-3, 1) and (-3, 1)

Problem 9 :

y = x2 + 5x + 8     {y = 2}

Solution :

y = x2 + 5x + 8

Applying y = 2, we get

2 = x2 + 5x + 8

x2 + 5x + 8 - 2 = 0

x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x = -2 and x = -3

The solutions are (-2, 2) and (-3, 2)

Problem 10 :

y = x2 - 5x + 1     {y = -3}

Solution :

y = x2 - 5x + 1

Applying y = -3, we get

-3 = x2 - 5x + 1

x2 - 5x + 1 + 3 = 0

x2 - 5x + 4 = 0

x2 - 1x - 4x + 4 = 0

x(x - 1) - 4(x - 1) = 0

(x - 1)(x - 4) = 0

x = 1 and x = 1

The solutions are (1, -3) and (1, -3)

Problem 11 :

f(x) = x2 + 4x + 11 find x when

i)  f(x) = 23     ii)  f(x) = 7

Solution :

f(x) = x2 + 4x + 11

i)  f(x) = 23

23 = x2 + 4x + 11

x2 + 4x + 11 - 23 = 0

x2 + 4x - 12 = 0

x2 + 6x - 2x - 12 = 0

x(x + 6) - 2(x + 6) = 0

(x + 6)(x - 2) = 0

x = -6 and x = 2

So, the solutions are (-6, 23) and (2, 23).

ii)  f(x) = 7

7 = x2 + 4x + 11

x2 + 4x + 11 - 7 = 0

x2 + 4x + 4 = 0

x2 + 2x + 2x + 4 = 0

x(x + 2) + 2(x + 2) = 0

(x + 2)(x + 2) = 0

x = -2 and x = -2

So, the solutions are (-2, 7) and (-2, 7).

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