Solve this following
Question:

Let $\mathrm{p}$ and $\mathrm{q}$ be two positive numbers such that $\mathrm{p}+\mathrm{q}=2$ and $\mathrm{p}^{4}+\mathrm{q}^{4}=272$. Then $\mathrm{p}$ and $\mathrm{q}$ are roots of the equation :

 

  1. $x^{2}-2 x+2=0$

  2. x^{2}-2 x+8=0

  3. $x^{2}-2 x+136=0$

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


Correct Option: , 4

Solution:

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

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

$16-16 p q+2 p^{2} q^{2}=272$

$(p q)^{2}-8 p q-128=0$

$(\mathrm{pq})^{2}-8 \mathrm{pq}-128=0$

$\mathrm{pq}=\frac{8 \pm 24}{2}=16,-8$

$\therefore \quad \mathrm{pq}=16$

$\therefore \quad$ Required equation : $x^{2}-(2) x+16=0$

 

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