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For $p, q \in R$, consider the real valued function $f ( x )=( x - p )^{2}- q , x \in R$ and $q >0$. Let $a _{1}, a _{2}, a _{3}$ and $a _{4}$ be in an arithmetic progression with mean $P$ and positive common difference. If $\left| f \left( a _{ i }\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f ( x )=0$ is.
$50$
$60$
$70$
$80$
Solution
$f(x)=0 \Rightarrow(x-p)^{2}-q=0$
Roots are $p+\sqrt{q}, p-\sqrt{q}$ absolute difference between roots $2 \sqrt{q}$.
Now, $\left|f\left(a_{i}\right)\right|=500$
Let $a_{1}, a_{2}, a_{3}, a_{4} a_{r} a_{1} a+d, a+2 d, a+3 d$
$\left|f\left(a_{4}\right)\right|=500$
$\left|\left(a_{1}-p\right)^{2}-q\right|=500$
$\Rightarrow\left(a_{1}-p\right)^{2}-q=500$
$\Rightarrow \frac{9}{4} d^{2}-q=500$
$\text { and }\left|f\left(a_{1}\right)\right|^{2}=\left|f\left(a_{2}\right)\right|^{2}$
$\left(\left(a_{1}-p\right)^{2}-q\right)^{2}=\left(\left(a_{2}-p\right)^{2}-q\right)^{2}$
$\left(\left(a_{1}-p\right)^{2}-\left(a_{2}-p\right)^{2}\right)\left(\left(a_{1}-p\right)^{2}-q+\left(a_{2}-p\right)^{2}-q\right)=0$
$\Rightarrow \frac{9}{4} d^{2}-q+\frac{d^{2}}{4}-q=0$
$2 q=\frac{10 d^{2}}{4} \Rightarrow q=\frac{5 d^{2}}{4}$
$\Rightarrow d^{2}=\frac{4 q}{5}$
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 $2 \sqrt{q}=2 \times \frac{50}{2}=50$
