Question #d4f12

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Question #d4f12

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Answer 1 · 2017-11-14T07:35:04

$\frac{49}{4}$ $49/4$

Explanation:

To make a complete square, we can make a square of ${(a x + b)}^{2}$ $(ax+b)^2$ .

${(a x + b)}^{2}$ $(ax+b)^2$
$= a^{2} x^{2} + 2 a b x + b^{2}$ $=a^2x^2+2abx+b^2$

In $x^{2} - 7 x$ $x^2-7x$ ,
$a^{2} = 1 and 2 a b = - 7$ $a^2=1 and 2ab=-7$
Therefore $a = 1, b = - \frac{7}{2}$ $a=1, b=-7/2$ or $a = - 1, b = \frac{7}{2}$ $a=-1, b=7/2$

Hence, the required value
$= b^{2}$ $=b^2$
$= {(- \frac{7}{2})}^{2} or {(\frac{7}{2})}^{2}$ $=(-7/2)^2 or (7/2)^2$
$= \frac{49}{4}$ $=49/4$

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Question #d4f12

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