Quadratic Equations Practice Questions — SAT — Page 4

Question 61Question →

The quadratic equation $x^2 + 10x + c = 7$ , where $c$ is a constant, has exactly one real solution. What is the value of $c$ ?

18

25

32

-18

Show answer & explanation

Answer: $32$

Answer

32

For the quadratic equation to have exactly one real solution, it must be written in the standard form

x^2 + 10x + (c - 7) = 0

, and its discriminant must equal zero. Setting the discriminant to zero gives

10^2 - 4(1)(c - 7) = 0

. Distributing the negative four yields

100 - 4c + 28 = 0

, which simplifies to

128 - 4c = 0

. Solving for the constant gives

c = 32

. Alternatively, the left side of the equation in standard form must be a perfect square trinomial, meaning

x^2 + 10x + 25 = 0

. Thus,

c - 7 = 25

, which yields

c = 32

.

Step-by-Step Solution

1

Subtract

7

from both sides of the equation to write it in the standard quadratic form

ax^2 + bx + c = 0

.

x^2 + 10x + (c - 7) = 0

To analyze the solutions of a quadratic equation using the discriminant, the equation must be in standard form.

2

Identify the coefficients and set the discriminant

b^2 - 4ac

to

0

since the equation has exactly one real solution.

10^2 - 4(1)(c - 7) = 0

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

3

Solve the equation

100 - 4(c - 7) = 0

for

c

by distributing the

-4

and combining like terms.

100 - 4c + 28 = 0 \implies 128 - 4c = 0 \implies 4c = 128 \implies c = 32

Isolating the variable

c

yields the value that makes the equation have exactly one real solution.

Key Concept

Discriminant of a quadratic equation

Alternative Method

Alternatively, you can complete the square. For the quadratic expression

x^2 + 10x + (c - 7)

to have exactly one real solution, it must be a perfect square trinomial of the form

(x + d)^2 = x^2 + 2dx + d^2

. Comparing coefficients,

2d = 10 \implies d = 5

, so the constant term must be

d^2 = 25

. Setting the constant term

c - 7 = 25

yields

c = 32

.

Estimated Time:1m 30s

Question 62Question →

The solutions to the quadratic equation $x^2 - 6x - 11 = 0$ can be written in the form $x = a \pm \sqrt{b}$ , where $a$ and $b$ are integers. What is the value of $a + b$ ?

8

17

23

43

Show answer & explanation

Answer: 23

Answer

23

Completing the square on the quadratic equation

x^2 - 6x - 11 = 0

yields the solutions

x = 3 \pm \sqrt{20}

. Comparing this to the given form

x = a \pm \sqrt{b}

identifies

a = 3

and

b = 20

. Summing these values gives

a + b = 23

.

Step-by-Step Solution

1

Add 11 to both sides of the equation to isolate the variable terms.

x^2 - 6x = 11

Preparing the quadratic equation to complete the square by separating constant terms.

2

Complete the square by adding the square of half the coefficient of

x

to both sides.

x^2 - 6x + 9 = 11 + 9

, which simplifies to

(x - 3)^2 = 20

Adding

(\frac{-6}{2})^2 = 9

creates a perfect square trinomial on the left side.

3

Take the square root of both sides and solve for

x

.

x - 3 = \pm\sqrt{20}

, which gives

x = 3 \pm \sqrt{20}

Taking the square root cancels the exponent and yields two possible solutions.

4

Compare the solutions to the given form

x = a \pm \sqrt{b}

to determine

a

and

b

, and calculate their sum.

a = 3

,

b = 20

, so

a + b = 3 + 20 = 23

Matching the template shows that

a

corresponds to

3

and

b

corresponds to

20

.

Key Concept

Solving quadratic equations by completing the square and matching solutions to a given form.

Alternative Method

Instead of completing the square, you can use the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

with

a = 1

,

b = -6

, and

c = -11

. This yields

x = \frac{6 \pm \sqrt{36 - 4(1)(-11)}}{2} = \frac{6 \pm \sqrt{80}}{2} = 3 \pm \frac{\sqrt{80}}{2} = 3 \pm \sqrt{20}

. Comparing this to the given form yields

a = 3

and

b = 20

, which sums to

23

.

Estimated Time:1m 30s

Question 63Question →

In the quadratic equation $2x^2 - 15x + c = 0$ , $c$ is a constant. If one of the solutions to the equation is $x = 6$ , what is the other solution?

Show answer & explanation

Answer: 1.5

Answer

1.5

The sum of the roots of a quadratic equation in the form

ax^2 + bx + c = 0

is given by

-\frac{b}{a}

. For the equation

2x^2 - 15x + c = 0

, the sum of the roots is

-\frac{-15}{2} = 7.5

. Since one of the solutions is

6

, the other solution is

7.5 - 6 = 1.5

. Alternatively, substituting

x = 6

into the equation yields

2(6)^2 - 15(6) + c = 0 \implies 72 - 90 + c = 0 \implies c = 18

. The equation becomes

2x^2 - 15x + 18 = 0

, which factors as

(2x - 3)(x - 6) = 0

. Setting the factor

2x - 3 = 0

gives the other solution,

x = 1.5

(or

\frac{3}{2}

).

Step-by-Step Solution

1

Find the sum of the roots of the quadratic equation using Vieta's formulas.

The sum of the roots is

-\frac{-15}{2} = 7.5

.

For any quadratic equation

ax^2 + bx + c = 0

, the sum of its roots is equal to

-\frac{b}{a}

.

2

Set up an equation using the known root to find the unknown root.

6 + x_2 = 7.5

, where

x_2

is the other root.

Since one root is

6

, adding it to the second root must equal the sum of the roots.

3

Solve for the second root.

x_2 = 1.5

.

Subtracting

6

from both sides of the equation yields

1.5

.

Key Concept

Sum of roots of a quadratic equation (Vieta's Formulas)

Estimated Time:1m 30s

Question 64Question →

In the quadratic equation $3x^2 - 18x + c = 0$ , $c$ is a constant. If the sum of the squares of the two real solutions to this equation is $26$ , what is the value of $c$ ?

Show answer & explanation

Answer: 15

Answer

The value of

c

is

15

.

To find the value of

c

, we apply Vieta's formulas to the equation

3x^2 - 18x + c = 0

. The sum of the solutions is

x_1 + x_2 = -\frac{-18}{3} = 6

, and the product of the solutions is

x_1 x_2 = \frac{c}{3}

. Using the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

, we substitute the given values:

26 = 6^2 - 2\left(\frac{c}{3}\right)

. This simplifies to

26 = 36 - \frac{2c}{3}

. Subtracting

36

from both sides yields

-10 = -\frac{2c}{3}

, which gives

2c = 30

, and thus

c = 15

. Alternatively, since the vertex of the corresponding parabola is at

x = -\frac{-18}{2(3)} = 3

, the two solutions can be represented symmetrically as

3 + d

and

3 - d

. The sum of their squares is

(3 + d)^2 + (3 - d)^2 = (9 + 6d + d^2) + (9 - 6d + d^2) = 18 + 2d^2

. Setting this equal to the given value of

26

yields

18 + 2d^2 = 26 \implies 2d^2 = 8 \implies d^2 = 4 \implies d = 2

. Therefore, the solutions are

3 + 2 = 5

and

3 - 2 = 1

. Substituting either solution back into the original equation, such as

x = 1

, yields

3(1)^2 - 18(1) + c = 0 \implies -15 + c = 0 \implies c = 15

.

Step-by-Step Solution

1

Find the sum and product of the solutions to the quadratic equation.

The sum of the solutions is

x_1 + x_2 = 6

, and their product is

x_1 x_2 = \frac{c}{3}

.

According to Vieta's formulas, for a quadratic equation

ax^2 + bx + c = 0

, the sum of the solutions is

-\frac{b}{a}

and the product of the solutions is

\frac{c}{a}

.

2

Relate the sum of the squares of the solutions to their sum and product using the algebraic identity.

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 \implies 26 = (6)^2 - 2\left(\frac{c}{3}\right)

.

The algebraic identity

(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2

can be rearranged to express the sum of squares

x_1^2 + x_2^2

in terms of the sum and product.

3

Solve the resulting linear equation for the constant

c

.

26 = 36 - \frac{2c}{3} \implies \frac{2c}{3} = 10 \implies c = 15

.

Subtracting

36

from both sides gives

-10 = -\frac{2c}{3}

. Multiplying by

-3

gives

30 = 2c

, which yields

c = 15

.

Key Concept

Vieta's Formulas and Algebraic Identities

Question 65Question →

A projectile is launched from a platform. The height $h(t)$ , in meters, of the projectile $t$ seconds after launch is modeled by the equation $h(t) = -5t^2 + v_0 t + h_0$ , where $v_0$ and $h_0$ are constants. If the projectile reaches its maximum height of $80$ meters at $3$ seconds after launch, what is the value of $h_0$ ?

5

35

80

125

Show answer & explanation

Answer: 35

Answer

The value of

h_0

is

35

, which represents the initial height of the projectile.

The maximum height of

80

meters at

3

seconds indicates that the vertex of the quadratic function is

(3, 80)

. Since the lead coefficient of the

t^2

term is

-5

, we can express the function in vertex form as

h(t) = -5(t - 3)^2 + 80

. To find the value of

h_0

, which represents the height at

t = 0

, we substitute

0

for

t

, yielding

h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35

. Thus, the initial height is

35

meters.

Step-by-Step Solution

1

Identify the coordinates of the vertex of the parabola from the context.

The maximum height is

80

meters at

3

seconds, so the vertex of the function is

(3, 80)

.

The vertex of a downward-opening parabola represents its maximum value.

2

Write the quadratic function in vertex form using the vertex

(3, 80)

and the given leading coefficient

a = -5

.

h(t) = -5(t - 3)^2 + 80

The vertex form of a quadratic equation is

h(t) = a(t - d)^2 + k

, where

(d, k)

is the vertex.

3

Evaluate the function at

t = 0

to find the initial height

h_0

.

h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35

The constant term

h_0

in standard form

h(t) = -5t^2 + v_0 t + h_0

represents the height when

t = 0

.

Key Concept

Vertex of a quadratic function and converting between vertex form and standard form.

Question 66Question →

In the quadratic equation $x^2 - kx + 5 = 0$ , $k$ is a constant. If the solutions to the equation are $x = 3 \pm \sqrt{a}$ , where $a$ is a positive constant, what is the value of $a$ ?

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. By relating the roots of a quadratic equation to its coefficients, the sum of the roots is

(3 + \sqrt{a}) + (3 - \sqrt{a}) = 6

, which equals the linear coefficient term

k

in

x^2 - kx + 5 = 0

. Using

k = 6

, we can rewrite the equation as

x^2 - 6x + 5 = 0

. Completing the square yields

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, or

x = 3 \pm \sqrt{4}

. Comparing this to the given expression

x = 3 \pm \sqrt{a}

gives

a = 4

. Alternatively, using the product of roots,

(3 + \sqrt{a})(3 - \sqrt{a}) = 9 - a

. Since the product of roots is the constant term 5, we have

9 - a = 5

, which directly yields

a = 4

.

Step-by-Step Solution

1

Find the sum of the given solutions to determine the value of the constant

k

.

k = 6

The sum of the solutions

(3 + \sqrt{a}) + (3 - \sqrt{a})

is

6

. For a quadratic equation in the form

x^2 - kx + c = 0

, the sum of the solutions is equal to

k

.

2

Substitute

k = 6

back into the quadratic equation.

x^2 - 6x + 5 = 0

This gives the complete quadratic equation with numerical coefficients.

3

Solve the quadratic equation by completing the square to express the solutions in the form

3 \pm \sqrt{a}

.

x = 3 \pm \sqrt{4}

Completing the square gives

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, so

x - 3 = \pm \sqrt{4}

.

4

Compare the obtained solutions to the given form

x = 3 \pm \sqrt{a}

to identify the value of

a

.

a = 4

Comparing

x = 3 \pm \sqrt{4}

with

x = 3 \pm \sqrt{a}

yields

a = 4

.

Key Concept

Relationship between roots and coefficients of a quadratic equation, or solving quadratic equations by completing the square.

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