Quadratic EquationMedium
Question
The sum of all the real solutions of the equation $\log_{(x + 3)}\left( 6x^{2} + 28x + 30 \right) = 5 - 2\log_{(6x + 10)}\left( x^{2} + 6x + 9 \right)$ is equal to :
Options
A.2
B.1
C.0
D.4
Solution
$\log_{x + 3}\lbrack(x + 3)(6x + 10)\rbrack = 5 - 2\log_{6x + 10}(x + 3)^{2}$
$$1 + \log_{x + 3}(6x + 10) = 5 - 4\log_{6x + 10}(x + 3) $$Let $\log_{(x + 3)}(6x + 10) = A$
$\Rightarrow A + \frac{4}{A} = 4$ or $A = 2$
$${\Rightarrow \log_{\text{(x+3)~}}(6x + 10) = 2 }{\Rightarrow 6x + 10 = (x + 3)^{2} }{\Rightarrow 6x + 10 = x^{2} + 9 + 6x }{\Rightarrow x^{2} = 1,x = \pm 1 }$$So sum of roots $= 0$
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