What value of $b$ $b$ makes the polynomial $x^{2} + b x + 16$ $x^2+ bx+16$ a perfect square?

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What value of $b$ $b$ makes the polynomial $x^{2} + b x + 16$ $x^2+ bx+16$ a perfect square?

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Answer 1 · 2016-08-03T11:33:27

$b = \pm 8$ $b=+-8$

Explanation:

$x^{2} + b x + 16$ $x^2+bx+16$ to become perfect square
$b^{2} - 4 a c = 0$ $b^2-4ac=0$ where $a = 1 and c = 16$ $a=1 and c=16$
Therefore
$b^{2} = 4 a c$ $b^2=4ac$
or
$b^{2} = 4 (16)$ $b^2=4(16)$
or
$b^{2} = 64$ $b^2=64$
or
$b = \pm \sqrt{64}$ $b=+-sqrt64$
or
$b = \pm 8$ $b=+-8$

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What value of $b$ $b$ makes the polynomial $x^{2} + b x + 16$ $x^2+ bx+16$ a perfect square?

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Science

Math

Humanities

... and beyond

What value of bb makes the polynomial x2+bx+16x^2+ bx+16 a perfect square?

1 Answer

Explanation:

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What value of $b$ $b$ makes the polynomial $x^{2} + b x + 16$ $x^2+ bx+16$ a perfect square?