a
^2 + 4b2 + 6a + 4b + 10 = 0, find
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[tex] Given \:a^{2}+4b^{2}+6a+4b+10=0 [/tex]
[tex] \implies a^{2}+6a+9 + 4b^{2}+4b+1 = 0 [/tex]
[tex] \implies [a^{2}+2\times a \times 3 + 3^{2} ]+[(2b)^{2}+2\times 2b\times 1 + 1^{2}] = 0 [/tex]
[tex] \implies ( a + 3 )^{2} + ( 2b + 1 )^{2} = 0 [/tex]
/* By Algebraic Identity */
[tex]\boxed{\pink{ x^{2} + 2xy + y^{2} = ( x + y )^{2} }}[/tex]
[tex] a + 3 = 0 \: and \: 2b + 1 = 0 [/tex]
[tex] \implies a = -3 \: and \: 2b = - 1 [/tex]
[tex] \implies a = -3 \: and \: b = \frac{- 1}{2} [/tex]
[tex] Now , \red{ Value \: of \: \frac{a}{b} } [/tex]
[tex] = \frac{(-3)}{\frac{-1}{2}} [/tex]
[tex] = (-3) \times \frac{2}{(-1)}[/tex]
[tex] = 6 [/tex]
Therefore.,
[tex] \red{ Value \: of \: \frac{a}{b} } \green { = 6}[/tex]
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