﻿ Solving Quadratic Equations Worksheet | Problems & Solutions

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1.
Find the two numbers whose sum is 30 and the product is maximum.
 a. 15, 15 b. 12, 18 c. 14, 16 d. 13, 17

#### Solution:

Let x represents one number.

Let 30 - x represents the other number.
[Sum is 30.]

Let y represents the product of the two numbers.

y = x(30 - x)
[Express as an equation.]

y = - x2 + 30x
[Multiply.]

y = - x2 + 30x is a quadratic with a = - 1 and b = 30.
[Compare with y = ax2 + bx + c.]

a = -1 < 0. So, the parabola opens downward with its vertex being the maximum point.

This maximum point occurs when x = - b2a.

x = - 30 / 2(-1) = 15
[Substitute and simplify.]

If x = 15, then 30 - x = 30 - 15 = 15.

So, the numbers are 15 and 15.

2.
Find two numbers whose difference is 12 and their product is minimum.
 a. 4, - 8 b. 6, - 6 c. - 9, 3 d. 7, - 5

#### Solution:

Let x represents one number.

Let 12 + x represents the other number.
[Since the difference is 12.]

Let y represents the product of the two numbers.

y = x(12 + x)
[Express as an equation.]

y = x2 + 12x
[Multiply.]

y = x2 + 12x is a quadratic with a = 1 and b = 12.
[Compare with y = ax2 + bx + c.]

a = 1 > 0. So, the parabola opens upward with its vertex being the minimum point.

This minimum point occurs when x = - b2a.

x = - 12 / 2(1)= - 6
[Substitute and simplify.]

If x = - 6, then 12 + x = 12 + (- 6) = 6.

So, the numbers are 6 and - 6.

3.
Check if the function $y$ = - 4$x$2 + 3 has a maximum or minimum value. Find that value.
 a. Maximum; 3 b. Minimum; 3 c. Maximum; - 3 d. Minimum; - 3

#### Solution:

y = - 4x2 + 3 is a quadratic function with a = - 4, b = 0 and c = 3.
[Compare with y = ax2 + bx + c.]

a = - 4 < 0. So, the parabola opens downward with its vertex being the maximum point.

This maximum point occurs when x = - b2a.

x = - 0 / 2(-4)= 0
[Substitute and simplify.]

Substitute this x - value in the given quadratic function to find the corresponding y - value.

y = - 4x2 + 3

= - (4)(0)2 + 3 = 3

For a quadratic function, the y - coordinate of the vertex is the maximum or minimum value of the function.

So, the given function has a maximum value 3.

4.
Sketch the graph of $y$ = $x$2 + 10$x$ + 25.

 a. Graph-4 b. Graph-1 c. Graph-2 d. Graph-3

#### Solution:

y = x2 + 10x + 25

a = 1, b = 10 and c = 25
[Compare with y = ax2 + bx + c.]

a = 1 > 0. So, the graph opens upward.

x - coordinate of the vertex = - b2a = - (10) / 2(1) = - 10 / 2= - 5

y - coordinate of the vertex = f( - 5) = (- 5)2 + 10(- 5) + 25 = 25 - 50 + 25 = 0

So, the vertex is the point (- 5, 0) and the axis of symmetry is the line x = - 5.

x2 + 10x + 25 = 0
[Substitute the values.]

x = - 5 (twice)
[Factor and simplify.]

c = 25, so, y - intercept = 25.

The graph just touches the x - axis at the point (- 5, 0).

The y - intercept (0, 25) is 5 units to the right of the axis of symmetry. So, its image point would be 5 units to the left of the axis. It is: (- 10, 25).

Graph the axis of symmetry. Plot all the points found and connect them with a smooth curve to get the graph of the given function.

5.
Find the equation of the axis of symmetry, the coordinates of the vertex, and the $x$ - and $y$ - intercepts for the function $y$ = - $x$2 + 6$x$.
 a. $x$ = 3; (3, 18); $x$ - intercept = 0 or 6; $y$ - intercept = 0 b. $x$ = 3; (3, 9); $x$ - intercept = 0 or 6; $y$ - intercept = 0 c. $x$ = - 3; (3, 13); $x$ - intercept = 0 or 6; $y$ - intercept = 0 d. $x$ = 3; (3, 9); $x$ - intercept = - 6; $y$ - intercept = 0

#### Solution:

y = - x2 + 6x is a quadratic with a = - 1, b = 6 and c = 0.
[Compare with y = ax2 + bx + c.]

The equation of the axis of symmetry is given by: x = - b2a

= - 6 / 2(-1)= 3
[Substitute and simplify.]

So, the equation of the axis of symmetry is x = 3.

The x - coordinate of the vertex of a quadratic function is given by: x = - b2a = 3

Substitute the x - coordinate of the vertex in the given quadratic function to find the corresponding y - coordinate.

y = - x2 + 6x

= - (3)2 + 6(3) = 9

So, the vertex of the given quadratic function is (3, 9).

Substitute y = 0 in the given quadratic function to find the x - intercept.

- x2 + 6x = 0

- x (x - 6) = 0
[Factor.]

Therefore, x = 0 or x = 6.
[Solve for x.]

c = 0, so, y - intercept = 0.

6.
Find the equation of the axis of symmetry, the coordinates of the vertex, and the $x$ - and $y$ - intercepts for the function $y$ = $x$2 + 12$x$ + 36.
 a. $x$ = 6; (6, 0); - 6; 36 b. $x$ = - 6; ( - 6, 72); - 6; 36 c. $x$ = - 6; ( - 6, 0); 6; 36 d. $x$ = - 6; ( - 6, 0); - 6; 36

#### Solution:

y = x2 + 12x + 36 is a quadratic with a = 1, b = 12 and c = 36.
[Compare with y = ax2 + bx + c.]

The equation of the axis of symmetry is given by x = - b2a

= - 12 / 2(1)= - 6
[Substitute and Simplify.]

So, the equation of the axis of symmetry is x = - 6.

The x - coordinate of the vertex of a quadratic function is given by x = - b2a = - 6

Substitute the x - coordinate of the vertex in the given quadratic function to find the corresponding y - coordinate.

y = x2 + 12x + 36

= (- 6)2 + 12( - 6) + 36 = 0

So, the vertex of the given quadratic function is (- 6, 0).

Substitute y = 0 in the given quadratic function to find the x - intercept.

x2 + 12x + 36 = 0

(x + 6)2 = 0
[Factor.]

Therefore, x = - 6 (twice).
[Solve for x.]

c = 36, so, y - intercept = 36.

7.
For what values of $a$ and $b$, the function $y$ = $a$$x$2 + $\mathrm{bx}$ + 6 will have its vertex at (- 6, -138)?
 a. $a$ = 4, $b$ = - 48 b. $a$ = 4, $b$ = 48 c. $a$ = - 4, $b$ = 48 d. $a$ = - 4, $b$ = - 48

#### Solution:

y = ax2 + bx + 6 is a quadratic with a = a, b = b and c = 6.
[Compare with y = ax2 + bx + c.]

The x - coordinate of the vertex of a quadratic function is given by: x = - b2a

- b2a = - 6

b = 12a
[Simplify.]

That is: 12a - b = 0 - - - - Equation (1)
[Express as an equation.]

Substitute the vertex (- 6, -138) in the given equation: y = ax2 + bx + 6.

-138 = a(- 6)2 + b(- 6) + 6

36a - 6b = -144
[Simplify.]

6a - b = -24 - - - - Equation (2)
[Divide throughout by 6.]

Solve equations (1) and (2) to get the values of a and b.

We get a = 4 and b = 48.

8.
Find the equation of the axis of symmetry, the coordinates of the vertex, and the $x$ - and $y$ - intercepts for the function $y$ = $x$2 + 2$x$ - 80.
 a. $x$ = - 1; ( - 1, -81); $x$ - intercept = - 10 or 8; $y$ - intercept = 80 b. $x$ = - 1; ( - 1, -81); $x$ - intercept = - 10 or 8; $y$ - intercept = - 80 c. $x$ = - 1; (1, -81); $x$ - intercept = - 10 or 8; $y$ - intercept = - 80 d. $x$ = - 1; ( - 1, 81); $x$ - intercept = - 10 or 8; $y$ - intercept = 80

#### Solution:

y = x2 + 2x - 80 is a quadratic with a = 1, b = 2 and c = - 80.
[Compare with y = ax2 + bx + c.]

The equation of the axis of symmetry is given by: x = - b2a

= - 2 / 2(1)= - 1
[Substitute and simplify.]

So, the equation of the axis of symmetry is x = - 1.

The x - coordinate of the vertex of a quadratic function is given by: x = - b2a = - 1

Substitute the x - coordinate of the vertex in the given quadratic function to find the corresponding y - coordinate.

y = x2 + 2x - 80

= (- 1)2 + 2( - 1) - 80 = -81

So, the vertex of the given quadratic function is (- 1, -81).

Substitute y = 0 in the given quadratic function to find the x - intercept.

x2 + 2x - 80 = 0

(x + 10)(x - 8) = 0
[Factor.]

Therefore, x = - 10, 8.
[Solve for x.]

c = -80, so, y - intercept = - 80.

9.
Find the equation of the axis of symmetry, the coordinates of the vertex, and the $x$ - and $y$ - intercepts for the function $y$ = + 3.
 a. $x$ = ($\frac{-1}{16}$); x - intercept = - 4 or 3; y - intercept = 12 b. $x$ = $\frac{7}{2}$ ; ($\frac{7}{2}$ , $\frac{-1}{16}$ ); x - intercept = 4 or 3; y - intercept = 12 c. $x$ = - $\frac{7}{8}$; x - intercept = 4 or 3; y - intercept = 12 d. None of the above

#### Solution:

y = 14x2-74 x + 3 is a quadratic with a = 1 / 4, b = - 7 / 4 and c = 3.
[Compare with y = ax2 + bx +c.]

The equation of the axis of symmetry is given by: x = - b2a

= - - 742 ×14 = 7 / 2
[Substitute and simplify.]

So, the equation of the axis of symmetry is x = 7 / 2 .

The x - coordinate of the vertex of a quadratic function is given by: x = - b2a = 7 / 2.

Substitute the x - coordinate of the vertex in the given quadratic function to find the corresponding y - coordinate.

y = 1 / 4x2 - 7 / 4x + 3

= 1 / 4(7 / 2)2 - 7 / 4(7 / 2) + 3 = -1 / 16.

So, the vertex of the given quadratic function is (7 / 2, -1 / 16).

Substitute y = 0 in the given quadratic function to find the x - intercept.

1 / 4x2 - 7 / 4x + 3 = 0

x2 - 7x + 12 = 0
[Multiply throughout by 4.]

(x - 4)(x - 3) = 0
[Factor.]

Therefore, x = 4, 3.
[Solve for x.]

c = 12, so, y - intercept = 12.

10.
Check if the function $y$ = - $x$2 + 12$x$ + 8 has a maximum or minimum value. Find that value.
 a. Maximum; 44 b. Maximum; 46 c. Minimum; 49 d. Minimum; 44

#### Solution:

y = - x2 + 12x + 8 is a quadratic with a = - 1, b = 12 and c = 8.
[Compare with y = ax2 + bx + c.]

a = - 1 < 0. So the parabola opens downward with its vertex being the maximum point.

This maximum point occurs when x = - b2a.

x = - 122(-1) = 6
[Substitute and simplify.]

Substitute this x - value in the given quadratic function to find the corresponding y - value.

y = - x2 + 12x + 8

= - (6)2 + 12(6) + 8 = 44

For a quadratic function, the y - coordinate of the vertex is the maximum or minimum value of the function.

So, the given function has a maximum value 44.