Question

if p and q are the zeros of polynomial f(x)=x^{2} -5x +k such that p-q = 1, find the value of k

Answer 1

Answer: k=2

Step-by-step explanation:

Given quadratix polynomial is x2 −6x + k

Since α, β are zeroes of the given polynomial, we have

Sum of roots (α+ β) = − (b/a) = − (−6/1) = 6

Prouduct of roots (αβ) =  (c/a) = (k/1) = k

Given (α+β)2+ 2αβ = 40

⇒ 62 + 2(k) = 40

⇒ 36 + 2k = 40

⇒ 2k = 40 − 36 = 4

∴ k = 2

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Answer 2

Answer:

The given polynomial: x² - 5x + k

This is a quadratic polynomial.

We know that, for a quadratic polynomial, the sum of zeros is given by:

→ Sum of Zeros ( Roots ) = -b/a

where, b is the coefficient of 'x' and a is the coefficient of 'x²'

According to the question,

• a = 1
• b = -5
• c = k

⇒ Sum of Zeros = p + q

⇒ p + q = = -(-5) / 1 = 5

Also it is given that p - q = 1

Hence solving the pair of linear equations we get:

⇒ p + q + p - q = 5 + 1   [ Addition of both the equations ]

⇒ 2p = 6

⇒ p = 6/2 = 3

Substituting in any of the equations we get:

⇒ 3 - q = 1

⇒ 3 - 1 = q

⇒ q = 2

Hence the value of 'p' is 3 and 'q' is 2.

Now substituting value of x as p in the given polynomial we get:

⇒ (p)² - 5(p) + k = 0  [Since p is a zero of the polynomial]

⇒ (3)² - 5(3) + k = 0

⇒ 9 - 15 + k = 0

⇒ -6 + k = 0

⇒ k = 6

Hence the value of k is 6.