If $$4x^2+9y^2=100$$
and $$(2x+3y)^2=150$$
then what is the value of 6xy?

$$(A)\ 5(2+\sqrt{6})$$
$$(B)\ 10\sqrt{6}$$
$$(C)\ 25$$
$$(D)\ 50$$
$$(E)\ 100$$

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

Hi LUANDATO,
Let's take a look at your question.

We will be solving this question using the following polynomial identity.
$$\left(a+b\right)^2=a^2+b^2+2ab$$

We are given with:
$$\left(2x+3y\right)^2=150$$

Evaluate the LHS using the polynomial identity (a+b)^2 = a^2 + b^2 + 2ab, we get:
$$\left(2x\right)^2+\left(3y\right)^2+2\left(2x\right)\left(3y\right)=150$$
$$4x^2+9y^2+2\left(6xy\right)=150...(i)$$

Also it is given in the question that:
$$4x^2+9y^2=100$$

Plugging in the value in eq(i), we get:
$$100+2\left(6xy\right)=150$$
$$2\left(6xy\right)=150-100$$
$$2\left(6xy\right)=50$$
$$\left(6xy\right)=\frac{50}{2}$$
$$6xy=25$$

Therefore value of 6xy is 25. Option C is correct.

Hope it helps.
I am available if you'd like any follow up.