Let f(x)=x^2+4x-31. For what value of a is there exaclty on real value of x such that f(x)=a? Thank you!

To find the value of a for which there is exactly one real value of x such that f(x) = a, we need to determine the discriminant of the quadratic equation f(x) = a.

The discriminant, denoted as Δ, is used to determine the nature of the roots of a quadratic equation. It is calculated using the formula Δ = b^2 - 4ac, where the quadratic equation is in the form ax^2 + bx + c = 0.

In this case, the given quadratic equation is f(x) = x^2 + 4x - 31, and we want to solve f(x) = a.

Setting f(x) equal to a, we get:

x^2 + 4x - 31 = a

Rearranging the equation, we obtain:

x^2 + 4x - (a + 31) = 0

Comparing this equation with the standard form ax^2 + bx + c = 0, we have:
a = 1,
b = 4, and
c = -(a + 31)

Now, we can calculate the discriminant. Substituting the values into the formula:

Δ = (4)^2 - 4(1)(-(a + 31))
= 16 + 4(a + 31)
= 16 + 4a + 124
= 4a + 140

For there to be exactly one real root, the discriminant should be equal to zero. Therefore:

4a + 140 = 0

Solving this equation for a, we subtract 140 from both sides:

4a = -140

Dividing by 4:

a = -35

Hence, the value of a for which there is exactly one real value of x such that f(x) = a is a = -35.