Problem of the Week

Problem 6 solution

Congratulations to Ivan at St Bede’s Inter-church School and Julia for solving Problem 6.

(i) By the quadratic formula, the roots are given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \text{.}$

So there are two roots if $b^2 > 4ac$ , there is one root if $b^2 = 4ac$ , and there are no roots if $b^2 < 4ac$ . Hence there is at most one root if and only if $b^2 \le 4ac$ .

(ii)

$\sum_{k=1}^n (a_kx+b_k)^2 \ge 0$

$\sum_{k=1}^n (a_k^2x^2+2a_kb_kx+b_k^2) \ge 0$

$\sum_{k=1}^n a_k^2x^2 + \sum_{k=1}^n 2a_kb_kx + \sum_{k=1}^n b_k^2 \ge 0$

$\left(\sum_{k=1}^n a_k^2\right)x^2 +\left(2\sum_{k=1}^n a_kb_k\right)x +\sum_{k=1}^n b_k^2 \ge 0$

If $\sum_{k=1}^n a_k^2 = 0$ , $a_k = 0$ for all $k$ and hence $\left(\sum_{k=1}^n a_k b_k\right)^2 = 0 \le \left(\sum_{k=1}^n a_k^2\right) \left(\sum_{k=1}^n b_k^2\right) = 0$ , as required. Otherwise, this is a quadratic in $x$ and has at most one root since it is non-negative for all $x$ . So by (i),