How to find the value of a and the value of b?

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How to find the value of a and the value of b?

$x^{2} - 16 x + a = {(x + b)}^{2}$ $x^2-16x+a=(x+b)^2$

2 Answers

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Answer 1 · 2017-03-13T07:18:24

$a = 64$ $a=64$ and $b = - 8$ $b=-8$

Explanation:

This appears to be a way of finding a number $a$ $a$ , which when added to $x^{2} - 16 x$ $x^2-16x$ results in a square of form ${(x + b)}^{2}$ $(x+b)^2$

We can write $x^{2} - 16 x + a = {(x + b)}^{2}$ $x^2-16x+a=(x+b)^2$ as

$x^{2} - 16 x + a = x^{2} + 2 b x + b^{2}$ $x^2-16x+a=x^2+2bx+b^2$

Now comparing coefficients of similar terms

$2 b = - 16$ $2b=-16$ or $b = - 8$ $b=-8$

and $a = b^{2} = {(- 8)}^{2} = 64$ $a=b^2=(-8)^2=64$

Answer 2 · 2017-03-13T07:20:25

$a = 64$ $a=64" "$ and $b = - 8$ $" "b=-8$

Explanation:

Given:

$x^{2} - 16 x + a = {(x + b)}^{2}$ $x^2-16x+a = (x+b)^2$

$x^{2} - 16 x + a = x^{2} + 2 b x + b^{2}$ $color(white)(x^2-16x+a) = x^2+2bx+b^2$

Equating the coefficients of $x$ $x$ , we find:

$- 16 = 2 b$ $-16 = 2b$

Hence:

$b = - 8$ $b = -8$

Then:

$b^{2} = {(- 8)}^{2} = 64$ $b^2 = (-8)^2 = 64$

So equating the constant terms, we find:

$a = 64$ $a = 64$

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How to find the value of a and the value of b?

$x^{2} - 16 x + a = {(x + b)}^{2}$ $x^2-16x+a=(x+b)^2$

2 Answers

Explanation:

Explanation:

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How to find the value of a and the value of b?

x^2-16x+a=(x+b)^2x2−16x+a=(x+b)2x^2-16x+a=(x+b)^2

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$x^{2} - 16 x + a = {(x + b)}^{2}$ $x^2-16x+a=(x+b)^2$