Question

Find value of k so that sum of zeroes of the polynomial 2x^2-(k+4)x+k is 3 times their product

Answer 1

Answer:-

Given Polynomial => 2x² - (k + 4)x + k

Let , a = 2 ; b = -(k + 4) ; c = k.

We know that,

Sum of the zeroes = - b/a

→ Sum = -[-(k + 4)]

→ Sum = k + 4

Product of the zeroes = c/a.

→ Product = k/2

And also given that,

Sum of the zeroes = 3(Product of the zeroes)

So,

→ k + 4 = 3(k/2)

→ k + 4 = 3k/2

→ 2(k + 4) = 3k

→ 2k + 8 = 3k

→ 3k - 2k = 8

→ k = 8

Hence, the value of k is 8.

Answer 2

Here, a = 2, b = -(k+4), c = k.

Sum of zeroes = -b/a

So, -b/a = -[-k+4]) / 2

-b/a = (k+4)/2

Product of zeroes = c/a

So, c/a = k/2

According to question :

3(Product of zeroes) = sum of zeroes

3(k/2) = (k+4) / 2

3k/2 = (k+4) / 2

3k = 2(k+4) / 2

3k = k+4

3k - k = 4

2k = 4

k = 4/2

k = 2

Hence, the value of k is 2.