If the quadratic equation x^2 + kx + 25 = 0 has exactly one real solution, and k > 0, what… — ACT

If the quadratic equation $x^2 + kx + 25 = 0$ has exactly one real solution, and $k > 0$ , what is the value of $k$ ?

Answer: 10

The value of

k

10

For any quadratic equation in standard form,

ax^2 + bx + c = 0

, the equation has exactly one real solution when the discriminant

b^2 - 4ac

is equal to

0

. Here, the equation is

x^2 + kx + 25 = 0

, so

a = 1

b = k

, and

c = 25

. Setting the discriminant to

0

gives

k^2 - 4(1)(25) = 0

, which simplifies to

k^2 - 100 = 0

. Solving for

k

yields

k = \pm 10

. Since we are given that

k > 0

, the only valid solution is

10

Set the discriminant of the quadratic equation to zero.

b^2 - 4ac = 0

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

Identify coefficients

a

b

, and

c

from the equation

x^2 + kx + 25 = 0

and substitute them into the discriminant equation.

k^2 - 4(1)(25) = 0

For

x^2 + kx + 25 = 0

, the coefficients are

a=1

b=k

, and

c=25

Simplify the equation and solve for

k

k^2 = 100

, so

k = 10

k = -10

Simplifying

k^2 - 100 = 0

gives

k^2 = 100

Apply the constraint

k > 0

to find the final value.

k = 10

The problem specifies that

k

must be greater than zero.

Using the discriminant (

b^2 - 4ac

) of a quadratic equation to determine when there is exactly one real solution.

Estimated Time:45s

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