Question

If p - q = 6 and p2?+ q2?= 116, what is the value of pq if (p?- q)2?= p2 ? 2pq + q2

Answer 1

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Answer 2

Answer:

The required answer for the value of 'pq' is found to be 40.

Step-by-step explanation:

We are given the identity of square of difference of two real numbers as:

$(p-q)^2=p^2+q^2-2pq$

Using this identity we have to find the value of 'pq', when we are given the values of $p-q=6$  and $p^2+q^2=116$.

Starting with the identity and substituting the value $p-q=6$ in the equation, we get:

$(p-q)^2=(6)^2=p^2+q^2-2pq$

Now, substituting the value of $p^2+q^2=116$, we get:

$(6)^2=p^2+q^2-2pq=116-2pq$

Simplifying this, we get:

$36=116-2pq$

Seperating the constant terms on one side of the sign of equality, we get:

$2pq=116-36$

Simplifying this, we get:

$2pq=80$

Dividing both sides by 2, we get:

$pq=40$

Thus, the correct value for 'pq' is found to be 40.