A^2 + 4b2 + 6a + 4b + 10 = 0, find /

1.	A^2 + 4b2 + 6a + 4b + 10 = 0, find /
Answer» $Given \:a^{2}+4b^{2}+6a+4b+10=0$ $\implies a^{2}+6a+9 + 4b^{2}+4b+1 = 0$ $\implies [a^{2}+2\times a \times 3 + 3^{2} ]+[(2b)^{2}+2\times 2b\times 1 + 1^{2}] = 0$ $\implies ( a + 3 )^{2} + ( 2b + 1 )^{2} = 0$ /* 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., $\red{ Value \: of \: \frac{a}{b} } \green { = 6}$ ••♪

Answer»

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