The quadratic equation x^2 + 10x + c = 7, where c is a constant, has exactly one real solu… — SAT

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

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

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.

Identify the coefficients and set the discriminant

b^2 - 4ac

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.

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.

Discriminant of a quadratic equation

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

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