Question

Show that (4p^2-3q)^2+48p^2q =(4p^2+3q^2)​

Answer 1

Given:

The statement is

`(4p^2-3q)^2+48p^2q=(4p^2+3q^2)`

To prove:

The given statement.

Solution:

We have,

`(4p^2-3q)^2+48p^2q=(4p^2+3q^2)`

Now,

`LHS=(4p^2-3q)^2+48p^2q`

Using
`(a-b)^2=a^2+b^2-2ab`, we get

`LHS=(4p^2)^2+(3q)^2-2(4p^2)(3q)+48p^2q`

`LHS=(4p^2)^2+(3q)^2-24p^2q+48p^2q`

`LHS=(4p^2)^2+(3q)^2+24p^2q`

It can be rewritten as

`LHS=(4p^2)^2+(3q)^2+2(4p^2)(3q)`

Using
`(a+b)^2=a^2+b^2+2ab`, we get

`LHS=(4p^2+3q^2)`

`LHS=RHS`

Hence proved.