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4-2.Quadratic Equations and Inequations
medium
A real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(a) ${\log _4}\left\{ {\,{{\log }_2}(\sqrt {x + 8} – \sqrt x )} \right\} = 0$
$ \Rightarrow {4^0} = {\log _2}\left( {\sqrt {x + 8} – \sqrt x } \right)$$ \Rightarrow {2^1} = \sqrt {x + 8} – \sqrt x $
$ \Rightarrow 4 = x + 8 + x – 2\sqrt {{x^2} + 8x} $
$ \Rightarrow 2\sqrt {{x^2} + 8x} = 2x + 4$ $ \Rightarrow {x^2} + 8x = {x^2} + 4 + 4x$
$ \Rightarrow 4x = 4$ $ \Rightarrow x = 1$.
Standard 11
Mathematics
