Question

If p+q+r=12 and p^2 +q^2+r^2=50, find pq+ qr +pr.​

Answer 1

Given:

• p + q + r = 12
• p² + q² + r² = 50

To Find:-

• pq + qr + pr

Understanding:-

In this question, we have given two different values which are p+q+r = 12 and p²+q²+r² = 50 and asked to find out value of pq+qr+pr. To solve these types of questions, we need some algebraic expressions which help us solve these easily.

Solution:

As we have mentioned above that we need algebraic expressions to solve,

We know that:-

• $\large{\boxed{\red{\small{\bf{(a+b+c)^2~=~a^2+b^2+c^2+2(ab+bc+ca)}}}}}$

Where:-

• p = a
• q = b
• r = c

Let's put value in formula :-

(p+q+r)² = p² + q² + r² +2(pq + qr + rs)

→ 12² = 50 + 2(pq + qr + pr)

→ 144 = 50 + 2(pq + qr + pr)

→ 144 - 50 = 2(pq + qr + pr)

→ 94 = 2(pq + qr + pr)

→ pq + qr + pr = 94/2

→ pq + qr + pr = 47

Therefore,

• $\large{\boxed{\red{\small{\bf{value~of~pq+qr+ pr~=~47}}}}}$

More algebraic identities related to this:-

• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² - 2ab
• (a+b)(a-b) = a² - b²
• (a + b + c)² = a^2+b^2+c^2+2(ab+bc+ca)
• (a+b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ - 3ab(a-b)