Statement I: lf $\alpha,\ \beta$ are the roots of $x^{2}-ax+b=0$, then the equation whose roots are $\displaystyle \frac{\alpha+\beta}{\alpha},\ \displaystyle \frac{\alpha+\beta}{\beta}$ is $bx^{2}-a^{2}x+a^{2}=0$ Statement II: lf $\alpha,\ \beta$ are the roots of $x^{2}-bx+c=0$ and $\alpha+h,\ \beta+h$ are the roots of $x^{2}+qx+r=0$, then $\mathrm{h}=\mathrm{b}-\mathrm{q}$. Which of the above statement(s) is(are) true.

For Statement I, Vieta's formulas give alpha + beta = a, so the new roots are a/alpha and a/beta, which sum to a(alpha + beta)/(alpha*beta) = a^2/b and have a product of a^2/(alpha*beta) = a^2/b. The quadratic equation with these roots is x^2 - (a^2/b)x + a^2/b = 0, which clears to bx^2 - a^2x + a^2 = 0, making Statement I true. For Statement II, the sum of roots gives alpha + beta = b for the first equation and (alpha + h) + (beta + h) = -q for the second. Substituting b into the second equation yields b + 2h = -q, so h = (-q - b)/2, which contradicts the claim that h = b - q, making Statement II false.

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