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 :

${p^2} + {q^2} = {(p + q)^2} - 2pq$<br><br>$= 4 - 2pq$<br><br>Now, ${\left( {{p^2} + {q^2}} \right)^2} = {p^4} + {q^4} + 2{p^2}{q^2}$<br><br>$\Rightarrow {\left( {4 - 2pq} \right)^2} = 272 + 2{p^2}{q^2}$<br><br>$\Rightarrow 16 + 4{p^2}{q^2} - 16pq = 272 + 2{p^2}{q^2}$<br><br>$\Rightarrow 2{p^2}{q^2} - 16pq - 256 = 0$<br><br>$\Rightarrow {p^2}{q^2} - 8pq - 128 = 0$<br><br>$\Rightarrow (pq - 16)(pq + 8) = 0$<br><br>$\Rightarrow pq = 16, - 8$ <br/><br/>Here, pq = - 8 is not possible as p and q are positive. <br/><br/>$\therefore$ pq = 16 <br/><br/>Now, the equation whose roots are p and q is <br><br>${x^2} - 2x + 16 = 0$