For p,qR, consider the real-valued function f(x)=(x-p)^2-q, xR, q0.

Question

For $p,q\in\mathbb{R}$, consider the real-valued function $f(x)=(x-p)^2-q$, $x\in\mathbb{R}$, $q>0$. Let $a_1,a_2,a_3,a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $|f(a_i)|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f(x)=0$ is \_\_\_\_\_\_\_\_.

Answer: 50

Accepted Answer

Write $a_1=p-3d, a_2=p-d, a_3=p+d, a_4=p+3d$ ($d>0$). Then $f(a_1)=f(a_4)=9d^2-q$, $f(a_2)=f(a_3)=d^2-q$.

Since $|9d^2-q|=|d^2-q|=500$ and $d
e0$: $9d^2-q = -(d^2-q) \Rightarrow 5d^2=q$. Then $9d^2-q=4d^2=500 \Rightarrow d^2=125 \Rightarrow q=625$.

Roots of $f(x)=0$ are $p\pm\sqrt q$, so the absolute difference is $2\sqrt{625}=50$.

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