Trigonometry
Roots of quadratic as tan/cot with angle sum condition
MJAT_TS2_P1
Grade 12
Question:
Let $P(x) = x^2 + ax + b$ and $Q(x) = x^2 + cx + d$ be quadratic polynomials with real coefficients. If $\tan\theta_1, \tan\theta_2$ are the roots of $P(x)$ and $\cot\theta_1, \cot\theta_2$ are the roots of $Q(x)$ for some $\theta_1, \theta_2 \in (0, \pi/2)$, such that $\theta_1 + \theta_2 = \pi/4$ and $P(1)\cdot Q(1) = 16$, then the value of $\dfrac{a}{c+d}$ is equal to:
Step-by-Step Solution
Key Concept: Since $\cot\theta_i = 1/\tan\theta_i$: Vieta's gives $\tan\theta_1+\tan\theta_2=-a$, $\tan\theta_1\tan\theta_2=b$, $\cot\theta_1+\cot\theta_2=-c = -a/b$, $\cot\theta_1\cot\theta_2=d=1/b$. So $bd=1$. From $\theta_1+\theta_2=\pi/4$: $\tan\theta_1+\tan\theta_2+\tan\theta_1\tan\theta_2=1 \Rightarrow -a+b=1$.
$b=4,a=3,c=3/4,d=1/4$. Then $\frac{a}{c+d} = \frac{3}{3/4+1/4} = \frac{3}{1} = 3$.
Correct Answer: C