Question

Solve : p2 + q2 = n2

Answer 1

$(p+q+n)(p+q-n)-2pq=0$

Step-by-step explanation:

We have,

$p^2+q^2=n^2$

To find, $p^2+q^2=n^2=?$

∴ $p^2+q^2=n^2$

Using algebraic identity,

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

⇒ $a^{2}+b^{2}=(a+b)^{2}-2ab$

⇒ $(p+q)^{2}-2pq=n^2$

⇒ $(p+q)^{2}-n^2-2pq=0$

⇒ $[(p+q)^{2}-n^2]-2pq=0$

Using algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

⇒ $(p+q)+n][(p+q)-n]-2pq=0$

⇒ $(p+q+n)(p+q-n)-2pq=0$

∴$(p+q+n)(p+q-n)-2pq=0$

Answer 2

Step-by-step explanation:

Given :

$p^2+q^2=n^2$

now, it can also be written as

$p^2+q^2 -n^2$= 0

Applying $a^2-b^2 = (a+b)(a-b)$

so the above equation can be written as

p^2+q^2 -n^2 = {(p+q)+n}{(p+q)-n}=0

⇒ either {(p+q)+n} =0

or {(p+q)-n}=0

therefore, n = p+q

or n= -(p+q)