Question

If the sum of the zeros of the quadratic polynomial $f(t) = kt^{2} +2t+3k$ is equal to their product, find the value of k.

Answer 1

SOLUTION :  

Given : The quadratic polynomial f(t) = kt² + 2t + 3k  

Let the two zeroes of the quadratic polynomial  f(t) = kt² + 2t + 3k  be α and β.

On comparing with at² + bt + c,

a = k , b= 2 , c= 3k

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -2/k

α + β = -2/k ………………..(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = 3k/k

αβ = 3 ………………………(2)

Now,

α + β = αβ

[sum of the zeroes of the quadratic polynomial is equal to their product]

−2/k = 3

3k= −2

k= -2/3

Hence, the value of k is −2/3.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

-2/3

Step-by-step explanation:

Let the zeroes of the quadratic polynomial be α,β.

Given Equation is kt² + 2t + 3k.

here, a = k,b = 2,c = 3k.

(i)

Sum of zeroes = -b/a

⇒ α + β = -2/k.

(ii)

Product of zeroes = c/a

⇒ αβ = 3k/k

⇒ αβ = 3

Given that Sum of the zeroes is equal to their product.

⇒ α + β = αβ

⇒ -2/k = 3

⇒ k = -2/3.

Therefore, the value of k = -2/3.

Hope it helps!