Let p and q be two real numbers such that p+q=3 and p^4+q^4=369.

Question

Let $p$ and $q$ be two real numbers such that $p+q=3$ and $p^4+q^4=369$. Then $\left(\dfrac1p+\dfrac1q\right)^{-2}$ is equal to \_\_\_\_

Answer: 4

Accepted Answer

Let $s=p+q=3,\ m=pq$. Then $p^2+q^2=s^2-2m=9-2m$, and:
$$p^4+q^4=(p^2+q^2)^2-2m^2=(9-2m)^2-2m^2=81-36m+2m^2=369.$$

$$2m^2-36m-288=0 \implies m^2-18m-144=0 \implies m=\frac{18\pm\sqrt{324+576}}2=\frac{18\pm30}2 = 24 \text{ or } -6.$$

For real $p,q$ we need $s^2-4m\ge0$: with $m=24$, $9-96<0$ (rejected); with $m=-6$, $9+24>0$ (valid). So $m=-6$.

$$\frac1p+\frac1q=\frac{p+q}{pq}=\frac{3}{-6}=-\frac12 \implies \left(\frac1p+\frac1q\right)^{-2}=\left(-\frac12\right)^{-2}=4.$$

Solutions · 1