Question

Q8)If α, β are zeroes of the polynomial 2x2 + 5x + k such that α2 + β2 + = 21 4 , then find k.

Answer 1

||✪✪ QUESTION ✪✪||

f α, β are zeroes of the polynomial 2x² + 5x + k such that α² + β² = 21/4 , then find k. ?

|| ★★ FORMULA USED ★★ ||

The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

→ Product of roots of the Equation is given by = c/a.

|| ✰✰ ANSWER ✰✰ ||

Given Equation is :- 2x² + 5x + k = 0

So, we have :-

→ a = 2

→ b = 5

→ c = k

So,

→ sum of Roots = α + β = (-b/a) = (-5/2) -----Equation(1)

→ Product of Roots = α * β = c/a = (k/2) ------Equation(2)

___________

Now, Given That :- α² + β² = 21/4

Using :- a² + b² = (a+b)² - 2ab

→ α² + β² = (α + β)² - 2α * β

→ 21/4 = (α + β)² - 2α * β

Putting both Values from Equation (1) & (2) now, we get,

→ 21/4 = (-5/2)² - 2*(k/2)

→ 21/4 = (25/4) - k

→ k = (25/4) - (21/4)

→ k = (25 - 21)/4

→ k = (4/4)

→ k = 1 . (Ans).

Hence, value of k will be 1.

Answer 2

QUESTION :-

If α, β are zeroes of the polynomial 2x² + 5x + k such that α² + β² = 21 4 , then find k.

SOLUTION :-

=> p(x) = 2x² +5x +k ------(given)

=> On comparing with this ax²+bx+c, then we get,

=> •a = 2 , •b = 5 , • c = k

So,

=> sum of zeroes = α+β

=> (-b/a)= (-5/2)

=> product of zeroes = αβ

=> c/a = (k/2)

Now,

=> we know that,

=> a²+b² = (a+b)²-2ab

=> .°. α²+β² = (α+β)²-2(αβ)

=> put values .

=> 21/4 = (-5/2)² - 2(k/2) (°.° α²+β² = 21/4)

=> 21/4 = 25/4 - k

=> K = (25-21)/4

=> K= 4/4

=> K=1.

therefore, the value of k is 1.