Question

a
^2 + 4b2 + 6a + 4b + 10 = 0, find
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Answer 1

$Given \:a^{2}+4b^{2}+6a+4b+10=0$

$\implies a^{2}+6a+9 + 4b^{2}+4b+1 = 0$

$\implies [a^{2}+2\times a \times 3 + 3^{2} ]+[(2b)^{2}+2\times 2b\times 1 + 1^{2}] = 0$

$\implies ( a + 3 )^{2} + ( 2b + 1 )^{2} = 0$

/* By Algebraic Identity */

$\boxed{\pink{ x^{2} + 2xy + y^{2} = ( x + y )^{2} }}$

$a + 3 = 0 \: and \: 2b + 1 = 0$

$\implies a = -3 \: and \: 2b = - 1$

$\implies a = -3 \: and \: b = \frac{- 1}{2}$

$Now , \red{ Value \: of \: \frac{a}{b} }$

$= \frac{(-3)}{\frac{-1}{2}}$

$= (-3) \times \frac{2}{(-1)}$

$= 6$

Therefore.,

$\red{ Value \: of \: \frac{a}{b} } \green { = 6}$

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