Let $p$ and $q$ be the roots of the quadratic equation $ax^2+bx+c=0$, $a\ne0$. If $a$, $b$ and $c$ are in arithmetic progression and $\frac{1}{p}+\frac{1}{q}=4$, then the value of…

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Let $p$ and $q$ be the roots of the quadratic equation $ax^2+bx+c=0$, $a\ne0$. If $a$, $b$ and $c$ are in arithmetic progression and $\frac{1}{p}+\frac{1}{q}=4$, then the value of $(p-q)^2$ is

Accepted Answer

D - $\dfrac{52}{81}$. 1/p+1/q = (p+q)/pq = -b/c = 4 so b = -4c. AP: 2b = a+c gives a = -9c. Then p+q = -b/a = -4/9, pq = c/a = -1/9. (p-q)^2 = 16/81 + 4/9 = 52/81.

Answer

A - $\dfrac{48}{81}$

Answer

B - $\dfrac{61}{81}$

Answer

C - $\dfrac{68}{81}$

Let \(p\) and \(q\) be the roots of the quadratic equation \(ax^2+bx+c=0\), \(a\ne0\). If \(a\), \(b\) and \(c\) are in arithmetic progression and \(\frac{1}{p}+\frac{1}{q}=4\), then the value of…

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