For p, q in R, consider the real valued funct | vedclass

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Question

$f(x)=0 \Rightarrow(x-p)^{2}-q=0$

Roots are $p+\sqrt{q}, p-\sqrt{q}$ absolute difference between roots $2 \sqrt{q}$.

Now, $\left|f\left(a_{i}\ri

Vedclass · Accepted Answer

$50$

Vedclass · Answer

$f(x)=0 \Rightarrow(x-p)^{2}-q=0$

Roots are $p+\sqrt{q}, p-\sqrt{q}$ absolute difference between roots $2 \sqrt{q}$.

Now, $\left|f\left(a_{i}ight)ight|=500$

Let $a_{1}, a_{2}, a_{3}, a_{4} a_{r} a_{1} a+d, a+2 d, a+3 d$

$\left|f\left(a_{4}ight)ight|=500$

$\left|\left(a_{1}-pight)^{2}-qight|=500$

$\Rightarrow\left(a_{1}-pight)^{2}-q=500$

$\Rightarrow \frac{9}{4} d^{2}-q=500$

$	ext { and }\left|f\left(a_{1}ight)ight|^{2}=\left|f\left(a_{2}ight)ight|^{2}$

$\left(\left(a_{1}-pight)^{2}-qight)^{2}=\left(\left(a_{2}-pight)^{2}-qight)^{2}$

$\left(\left(a_{1}-pight)^{2}-\left(a_{2}-pight)^{2}ight)\left(\left(a_{1}-pight)^{2}-q+\left(a_{2}-pight)^{2}-qight)=0$

$\Rightarrow \frac{9}{4} d^{2}-q+\frac{d^{2}}{4}-q=0$

$2 q=\frac{10 d^{2}}{4} \Rightarrow q=\frac{5 d^{2}}{4}$

$\Rightarrow d^{2}=\frac{4 q}{5}$

From equation $(1)$ $\frac{9}{4} \cdot \frac{4 \cdot q}{5}-q=500$

$\frac{4 q}{5}=500$

$\frac{4 q}{5}=500$

and $2 \sqrt{q}=2 	imes \frac{50}{2}=50$

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