Find the value of $a$ and $b$ in $x^{2} - 16 x + a = {(x + b)}^{2}$

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Solution

Compare coefficients in $x^{2} - 16 x + a = {(x + b)}^{2}$ to find value of $a$ and $b$
$x^{2} - 16 x + a = {(x + b)}^{2}$
We will expand ${(x + b)}^{2}$ using the identity
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
We get the equation as
$x^{2} - 16 x + a = x^{2} + 2 b x + b^{2}$
On equating the coefficients on L.H.S and R.H.S
$- 16 = 2 b$
$b = - \frac{16}{2}$
$b = - 8$
and
$a = b^{2} a = {(- 8)}^{2} a = 64$
Hence $a = 64$ and $b = - 8$

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