Which value of b would make $16 x^{2} - b x + 25$ $16x^2 -bx+25$ a perfect square trinomial?

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Which value of b would make $16 x^{2} - b x + 25$ $16x^2 -bx+25$ a perfect square trinomial?

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Answer 1 · 2017-01-11T04:39:21

b = 40 and -40

Explanation:

General form of Perfect square trinomial is $a^{2} + 2 a b + b^{2}$ $a^2+2ab+b^2$

Therefore from
$16 x^{2} - b x + 25$ $16x^2-bx+25$

$a^{2} = \sqrt{16 x^{2}}, b^{2} = 25$ $a^2=sqrt(16x^2), b^2=25$ , then

$a = \pm 4 x, b = \pm 5$ $a = +-4x, b=+-5$

take consideration a=4x and b=-5 (different sign), then
$- b x = 2 (4 x) (- 5)$ $-bx = 2(4x)(-5)$
$- b x = - 40 x$ $-bx = -40x$
$b = 40$ $b=40$
The perfect square is ${(4 x - 5)}^{2} = 16 x^{2} - 40 x + 25$ $(4x-5)^2 = 16x^2-40x+25$ .

if we consider a=4x and b=5 (same sign), then
$- b x = 2 (4 x) (5)$ $-bx = 2(4x)(5)$
$- b x = 40 x$ $-bx = 40x$
$b = - 40$ $b=-40$
The perfect square is ${(4 x + 5)}^{2} = 16 x^{2} + 40 x + 25$ $(4x+5)^2=16x^2+40x+25$ .

The first solution ${(4 x - 5)}^{2}$ $(4x-5)^2$ is the best solution after comparing the expression given.

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Which value of b would make $16 x^{2} - b x + 25$ $16x^2 -bx+25$ a perfect square trinomial?

1 Answer

Explanation:

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Which value of b would make $16 x^{2} - b x + 25$ $16x^2 -bx+25$ a perfect square trinomial?