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 = \frac{-p ± \sqrt{p^2-4q}}{2},

i.e. x=−p+p2−4q2or−p−p2−4q2x =\frac{-p + \sqrt{p^2-4q}}{2} or \frac{-p - \sqrt{p^2-4q}}{2}.

If these two roots differ by 1, then

−p+p2−4q2−−p−p2−4q2=1\frac{-p + \sqrt{p^2-4q}}{2} - \frac{-p - \sqrt{p^2-4q}}{2} = 1

i.e. −p+p2−4q2+p+p2−4q2=2p2−4q2=p2−4q=1\frac{-p + \sqrt{p^2-4q}}{2} + \frac{p + \sqrt{p^2-4q}}{2}= \frac{2\sqrt{p^2-4q}}{2} =\sqrt{p^2-4q}=1.

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).