Question

If p²+q²=4pq then prove that prove that p⁴+q⁴=14p²q²

Answer 1

Answer:

Given

p² + q²= 4pq [squaring both side]

=> (p² + q²)² = (4pq)²

=> (p²)² + 2p²q² + (q²)² = 16p²q²

=> p⁴+ q⁴ = 16p²q² – 2p²q²

=> p⁴+ q⁴= 14p²q²

hence proved

Answer 2

Given,

$p^{2} +q^{2} =4pq$-----(1)

We have to prove the equation,

$p^{4} +q^{4} =14p^{2} q^{2}$------(2)

In order to prove the above equation 2, we have to use the first equation,

$p^{2} +q^{2} =4pq$

Squaring on both sides,

We get

$(p^{2} +q^{2} )^{2} =(4pq)^{2}$

Use the formula $(a+b)^{2} =a^{2} +b^{2} +2ab$

Here $a=p^{2} ,b=q^{2}$

Substitute these values in the formula.

We get,

$(p^{2}) ^{2} +(q^{2}) ^{2}+2p^{2} q^{2} =16p^{2} q^{2}$

$p^{4} +q^{4} =16p^{2} q^{2} -2p^{2} q^{2}$

$p^{4} +q^{4} =14p^{2} q^{2}$

Hence proved.

Therefore, equation 2 is proved from the equation 1.