Quadratics Difficulty 4

TMUA practice: Quadratics, problem 2

A TMUA-calibre quadratics problem at difficulty 4 of 5, with the full worked solution.

The question

Consider the quadratic equation

x^{2}+bx+c=0

. The number of pairs

(b,c)

for which the equation has solutions of the form

\cos\alpha

and

\sin\alpha

for some

\alpha

0

1

2

\text{infinite}

The correct answer is highlighted. D

Worked solution

If the roots are $\cos\alpha$ and $\sin\alpha$ , then by Vieta’s formulas

$b=-(\cos\alpha+\sin\alpha), \qquad c=\cos\alpha\sin\alpha.$

A relation between $b$ and $c$ . Squaring the sum,

b^{2}=(\cos\alpha+\sin\alpha)^{2}=\cos^{2}\alpha+\sin^{2}\alpha+2\sin\alpha\cos\alpha=1+2c.

So every admissible pair $(b,c)$ lies on the parabola $b^{2}=1+2c$ .

Infinitely many pairs. As $\alpha$ varies continuously over $[0,2\pi)$ , the pair $\bigl(b,c\bigr)=\bigl(-(\cos\alpha+\sin\alpha),\ \cos\alpha\sin\alpha\bigr)$ varies continuously and takes infinitely many distinct values (for example $c=\tfrac12\sin2\alpha$ alone already takes infinitely many values). So there are infinitely many such pairs.

Answer: D.