Solve the following quadratic equation: 4^ (x+1) +4^ (1-x) =10 - Vidyadip

Question

Solve the following quadratic equation:
$ 4^{(x+1)}+4^{(1-x)}=10 $

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Answer

$ 4^{(x+1)}+4^{(1-x)}=10 $
$ 4^x \cdot 4^1+4^1 \cdot 4^{-x}=10 $
$ \Rightarrow 4 y+\frac{4}{y}=10 \text { where } 4^x=y $
$ \Rightarrow 4 y^2-10 y+4=0 $
$ \Rightarrow 4 y^2-8 y-2 y+4=0 $
$ \Rightarrow 4 y(y-2)-2(y-2)=0 $
$ \Rightarrow(y-2)(4 y-2)=0 $
$ \Rightarrow y-2=0 \text { or } 4 y-2=0 $
$ \Rightarrow y=2 \text { or } y=\frac{2}{4}=\frac{1}{2} $
$ \Rightarrow y=2 \text { or } y=\frac{1}{2}$
In case I:
$ \Rightarrow 4^x=2 $
$ \Rightarrow(2)^{2 x}=(2)^1 $
$ \Rightarrow 2 x=1 $
$ \Rightarrow x=\frac{1}{2}$
In case II:
$\Rightarrow4^\text{x}=\frac{1}{2}$
$\Rightarrow(2)^\text{2x}=\Big(\frac{1}{2}\Big)^1$
$\Rightarrow(2)^{\text{2x}}=(2)^{-1}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence, $\frac{1}{2},-\frac{1}2{}$ are the roots of given equation.

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