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

Sagot :

Given:

The statement is

[tex](4p^2-3q)^2+48p^2q=(4p^2+3q^2)[/tex]

To prove:

The given statement.

Solution:

We have,

[tex](4p^2-3q)^2+48p^2q=(4p^2+3q^2)[/tex]

Now,

[tex]LHS=(4p^2-3q)^2+48p^2q[/tex]

Using [tex](a-b)^2=a^2+b^2-2ab[/tex], we get

[tex]LHS=(4p^2)^2+(3q)^2-2(4p^2)(3q)+48p^2q[/tex]

[tex]LHS=(4p^2)^2+(3q)^2-24p^2q+48p^2q[/tex]

[tex]LHS=(4p^2)^2+(3q)^2+24p^2q[/tex]

It can be rewritten as

[tex]LHS=(4p^2)^2+(3q)^2+2(4p^2)(3q)[/tex]

Using [tex](a+b)^2=a^2+b^2+2ab[/tex], we get

[tex]LHS=(4p^2+3q^2)[/tex]

[tex]LHS=RHS[/tex]

Hence proved.