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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0,

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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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Re: If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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If roots are real and equal then it must be
Discrimanate i.e
b^2 - 4ac = 0
By solving we get b = 80
Hence d...

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Re: If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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New post 16 Apr 2018, 14:16
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


Quadratic equation takes only one X when it looks like (x-n)^2 or (x+n)^2. In our case we have (ax-n)^2 => (ax)^2-2anx+n^2 => bx = 2*5*8*x = 80x. Answer (D)
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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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New post 20 Apr 2018, 11:25
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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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0,   [#permalink] 20 Apr 2018, 11:25
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