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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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16 Apr 2018, 05:13
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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 OA will be automatically revealed on Monday 23rd of April 2018 05:13:16 AM Pacific Time Zone
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Posts: 63
Re: If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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16 Apr 2018, 11:33
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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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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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Joined: 22 May 2016
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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, [#permalink]

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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
$$b^2 - 4ac = 0$$
$$a = 25$$
$$c = 64$$

$$b^2 - (4)(25)(64) = 0$$
$$b^2 - 6400 = 0$$
$$b^2 = 6400$$
$$b = 80$$

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.

$$b = 80$$

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

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