Question

If p and q are zeros of polynomial x^2 - 10x +k, such that p^2 + q^2 = 52. Find the value of k​

Answer 1

Answer.

• The value of k is 24.

Solution.

Given polynomial:

→ f(x) = x² - 10x + k

Comparing f(x) with ax² + bx + c, we get,

→ a = 1

→ b = -10

→ c = k

As p and q are the zeros of f(x),

→ p + q = -b/a

           =  -(-10)/1

           = 10

→ pq = c/a

        = k/1

        = k

It's given that,

→ p² + q² = 52

→ p² + q² + 2pq - 2pq = 52  (Add and subtract)

→ (p + q)² - 2pq = 52

Substituting the necessary values, we get,

→ 10² - 2k = 52

→ 100 - 52 = 2k

→ 2k = 48

→ k = 24.

★ So, the value of k in this polynomial is 24.

Required Polynomial:

→ f(x) = x² - 10x + 24

Answer 2

k = 24

Explanation :

Given , p,q are zeros

x² - 10x + k

p² + q² = 52

comparing , ax²+bx+c

a = 1 , b = -10 , c = k

p+q = -b/a = -(-10)/1 = 10

pq = c/a = k/1 = k

we know that , p²+q² - 2pq = 52

such that ,

(p+q)² - 2pq = 52

(10)² - 2k = 52

100 - 2k = 52

2k = 100-52 = 48

k = 24

The value of k is 24

hope this will help u