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Question

If \(\alpha, \beta\) are the roots of \(2x^2-3x+4=0\), then the equation whose roots are \(\alpha^2\) and \(\beta^2\) is

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Solution

Let \(f(x)=2x^2-3x+4=0 \rightarrow\) roots are \(\alpha, \beta\)
The equation whose roots are \(\alpha^2, \beta^2\) is \(f(\sqrt{x})=0\)
\(\Rightarrow 2x-3\sqrt{x}+4=0\\
\Rightarrow 2x+4=3\sqrt{x}\)
\(\Rightarrow (2x+4)^2=9x \\
\Rightarrow 4x^2+7x+16=0\)

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