Prove that $a x^{2} + 2 b x y + c y^{2} + 2 d x + 2 e y + f$ is a perfect square, if $b^{2} = a c$ , $d^{2} = a f$ , $e^{2} = c f$ .

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Solution

${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$

As for the equation $a x^{2} + 2 b x y + c y^{2} + 2 d x + 2 e y + f$

$b^{2} = a c$
$d^{2} = a f$
$e^{2} = c f$

Now put the value of this in the above equation we get ,

$a x^{2} + 2 \sqrt{a c} x y + c y^{2} + 2 \sqrt{a f} x + 2 \sqrt{c f} y + f$

${(\sqrt{a})}^{2} x^{2} + {(\sqrt{c})}^{2} y^{2} + {(\sqrt{f})}^{2} + 2 \sqrt{a c} x y + 2 \sqrt{a f} x + 2 \sqrt{c f} y$

Thus this can be written as ${(a + b + c)}^{2}$

Therefore, ${(\sqrt{a} x + \sqrt{c} y + \sqrt{f})}^{2}$

Hence this is the perfect square.

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