Bunuel wrote:

If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, where b > 0, what is the value of b?

A. 26

B. 40

C. 52

D. 80

E. 100

The expression \(b^2 - 4ac\) is the "discriminant" of the quadratic equation.

If \(b^2 - 4ac > 0\), there are two solutions

If \(b^2 - 4ac = 0\), there is one solution

If \(b^2 - 4ac < 0\), there are no solutions

There is one solution here, so

\(b^2 - 4ac = 0\)

\(a = 25\)

\(c = 64\)

\(b^2 - (4)(25)(64) = 0\)

\(b^2 - 6400 = 0\)

\(b^2 = 6400\)

\(b = 80\)Answer D

Alternatively, try factoring the quadratic equation. True, we don't know \(b\), but the first and third terms are big hints.

Notice the coefficient of \(ax^2\) and the constant, \(c\): 25 and 64 are perfect squares

There is one solution. Some root times itself = this quadratic equation

The algebraic identity \((x-y)^2\) has such a factor pattern (as does \((x+y)^2\)).

That is

\((x-y)^2=(x-y)(x-y)=x^2-2xy+ y^2\)

Knowing the identity is not a must here, but it helps. Either way:

Use the square roots of the first and third terms to factor\((5x-8)(5x-8)=25x^2-80x+64\)

That works. One solution.

\(b = 80\)

Answer D

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-- Albert Camus, "Return to Tipasa"