2230.63 – Almost Equal Roots

Consider the equation x2+px+q=0x^2 + px + q = 0, where pp and qq are positive numbers. (Math problems are always asking you to "consider" something.) If the roots of this equation differ by 1, then pp equals:

  1. 4q+1\sqrt{4q+1}

  2. q−1q-1

  3. −4q+1-\sqrt{4q+1}

  4. q+1q+1

  5. 4q−1\sqrt{4q-1}

Solution

Here we go: x2+px+q=0→x=−p±p2−4q2x^2 + px + q = 0 \rightarrow x = \dfrac{-p \pm \sqrt{p^2-4q}}{2}. If these two roots differ by 1, then

1=−p+p2−4q2−−p−p2−4q2=−p+p2−4q2+p+p2−4q2=2p2−4q2=p2−4q\begin{align*}1 &= \dfrac{-p + \sqrt{p^2-4q}}{2} - \dfrac{-p - \sqrt{p^2-4q}}{2} \\[1em]&= \dfrac{-p + \sqrt{p^2-4q}}{2} + \dfrac{p + \sqrt{p^2-4q}}{2} \\[1em]&= \dfrac{2\sqrt{p^2-4q}}{2} \\[1em]&= \sqrt{p^2-4q}\end{align*}

So p2−4q=1→p2−4q=1→p2=4q+1→p=4q+1\sqrt{p^2-4q} = 1 \rightarrow p^2-4q = 1 \rightarrow p^2 = 4q+1 \rightarrow p = \sqrt{4q+1} (pp was given as positive.)

So it's (a).

You might verify this by building a quadratic that fits, e.g. (x+2)(x+3)(x+2)(x+3).