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,

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

⇒ $(p+q)+n][(p+q)-n]-2pq=0$

⇒ $(p+q+n)(p+q-n)-2pq=0$

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

Answer 2

Thus the solution can be written as n = - (p+q)

Step-by-step explanation:

Given data:

p^2 + q^2 = n^2

Now, it can also be written as = 0

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

The above equation can be written as

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

⇒ {(p+q)+n} =0

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

Therefore, n = p+q

or n = - (p+q)