Q161 / 270 · NDA 2022 (Phase 2) · Mathematics

Let $p, q(p \gt q)$ be the roots of the quadratic equation $x^2+b x+c=0$ where $c \gt 0$. If $p^2+q^2-11 p q=0$, then what is $p-q$ equal to?

  • Correct answer
    A
    $3 \sqrt{c}$
  • B
    $3 c$
  • C
    $9 \sqrt{c}$
  • D
    $9 c$

Solution

Given that roots of $x^2+b x+c=0$ are $p$ and $q$. $\therefore \quad p+q=-b \text { and } p q=c$ Given that $\begin{aligned} & p^2+q^2-11 p q=0 \\ & \Rightarrow(p+q)^2-13 p q=0 \\ & \Rightarrow b^2-13 c=0 \end{aligned}$ $\Rightarrow b^2-13 c$ ...(i) $\begin{aligned} & \text { Now }(p-q)^2=(p+q)^2-4 p q \\ & =b^2-4 c=13 c-4 c=9 c \\ & p-q=3 \sqrt{c} \end{aligned}$ [from (i)]

Hence, the correct answer is option: 'A'.

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