Let p and q be two positive numbers such that p + q = 2 and p4+q4 = 272.

Question

Let p and q be two positive numbers such that p + q = 2 and p4+q4 = 272. Then p and q are
roots of the equation :

Accepted Answer

${p^2} + {q^2} = {(p + q)^2} - 2pq$

$= 4 - 2pq$

Now, $${\left( {{p^2} + {q^2}} \right)^2} = {p^4} + {q^4} + 2{p^2}{q^2}$$

$$ \Rightarrow {\left( {4 - 2pq} \right)^2} = 272 + 2{p^2}{q^2}$$

$$ \Rightarrow 16 + 4{p^2}{q^2} - 16pq = 272 + 2{p^2}{q^2}$$

$$ \Rightarrow 2{p^2}{q^2} - 16pq - 256 = 0$$

$$ \Rightarrow {p^2}{q^2} - 8pq - 128 = 0$$

$$ \Rightarrow (pq - 16)(pq + 8) = 0$$

$$ \Rightarrow pq = 16, - 8$$

Here, pq = - 8 is not possible as p and q are positive.

$$ \therefore $$ pq = 16

Now, the equation whose roots are p and q is

${x^2} - 2x + 16 = 0$

Solutions · 1