Let $f: R \rightarrow R$ be defined by $f(x)=2 x+3$. If $\alpha$, $\beta$ are the roots of the equation $f\left(x^2\right)-2 f\left(\frac{x}{2}\right)-1=0$, then $\alpha^2+\beta^2=$

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Aicharya · Accepted Answer

$\begin{array}{rlrl} & & f(x)=2 x+3 \ & & f\left(x^2ight)-2 f\left(\frac{x}{2}ight)-1=0 \ \Rightarrow & 2 x^2+3-2(x+3)-1=0 \ \Rightarrow & & 2 x^2-2 x-4=0 \ \Rightarrow & & x^2-x-2=0 \ \Rightarrow & & x=2 	ext { or } x=-1 \ 	herefore & & \alpha^2+\beta^2=4+1=5\end{array}$

Let $f: R \rightarrow R$ be defined by $f(x)=2 x+3$. If $\alpha$, $\beta$ are the roots of the equation…

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