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(B) Given the quadratic equation $2x^2 - 3x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = -(-3)/2 = 3/2$ and the product of roots $\al

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$4x^2 + 7x + 16 = 0$

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(B) Given the quadratic equation $2x^2 - 3x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = -(-3)/2 = 3/2$ and the product of roots $\alpha\beta = 4/2 = 2$.We need to find the equation whose roots are $\alpha^2$ and $\beta^2$.The sum of the new roots is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (3/2)^2 - 2(2) = 9/4 - 4 = (9 - 16)/4 = -7/4$.The product of the new roots is $\alpha^2\beta^2 = (\alpha\beta)^2 = (2)^2 = 4$.The required quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (-7/4)x + 4 = 0$.$x^2 + 7/4x + 4 = 0$.Multiplying the entire equation by $4$,we get $4x^2 + 7x + 16 = 0$.

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

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