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(A) The roots of $f(x) = (x - p)^2 - q = 0$ are $x = p \pm \sqrt{q}$.The absolute difference between the roots is $|(p + \sqrt{q}) - (p - \sqrt{q})| =

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) The roots of $f(x) = (x - p)^2 - q = 0$ are $x = p \pm \sqrt{q}$.The absolute difference between the roots is $|(p + \sqrt{q}) - (p - \sqrt{q})| = 2\sqrt{q}$.Given $a_1, a_2, a_3, a_4$ are in arithmetic progression with mean $p$ and common difference $d > 0$,we have $a_1 = p - \frac{3d}{2}, a_2 = p - \frac{d}{2}, a_3 = p + \frac{d}{2}, a_4 = p + \frac{3d}{2}$.Since $|f(a_i)| = 500$,we have $|(a_i - p)^2 - q| = 500$.For $i=4$,$|(\frac{3d}{2})^2 - q| = 500 \Rightarrow |\frac{9d^2}{4} - q| = 500$.For $i=1$,$|(-\frac{3d}{2})^2 - q| = 500 \Rightarrow |\frac{9d^2}{4} - q| = 500$.For $i=2$,$|(-\frac{d}{2})^2 - q| = 500 \Rightarrow |\frac{d^2}{4} - q| = 500$.Since $q > 0$ and $d > 0$,$\frac{9d^2}{4} - q = 500$ and $q - \frac{d^2}{4} = 500$ (as $\frac{d^2}{4} Adding these equations: $(\frac{9d^2}{4} - q) + (q - \frac{d^2}{4}) = 500 + 500$ $\Rightarrow 2d^2 = 1000$ $\Rightarrow d^2 = 500$.Substituting $d^2 = 500$ into $q - \frac{d^2}{4} = 500$,we get $q - 125 = 500 \Rightarrow q = 625$.The absolute difference between the roots is $2\sqrt{q} = 2\sqrt{625} = 2 	imes 25 = 50$.

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