Question

if the sum of the zeros of the polynomial kx^2+6x-3k is equal to twice their product find k​

Answer 1

ANSWER:

• Value of k is 1.

GIVEN:

• P(x) = kx²+6x-3k
• Sum of zeros of the given polynomial = 2(Product of zeros )

SOLUTION:

Formula:

Sum of zeros(α+β) = -(coefficient of x)/(coefficient of x²)

Product of zeros(αβ) = Constant term/Coefficient of x²

P(x) = kx²+6x-3k

Here:

Sum of zeros (α+β)= -6/k

Product of zeros (α β)= -3k/k

Product of zeros (α β) = -3

=>Sum of zeros of the given polynomial = 2(Product of zeros )

=> (α+β) = 2(αβ)

=> -6/k = 2(-3)

=> -6/k = -6

=> k = (-6)/(-6)

=> k = 1

Value of k is 1.

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab

Answer 2

GIVEN:

Quadratic polynomial = kx² + 6x - 3k

Sum of zeroes of polynomial is twice to their product.

TO FIND:

The value of k.

SOLUTION:

We know that,

Sum of zeroes = (α+β) = -b/a

Product of zeroes = αβ = c/a

From the above polynomial,

a = k ; b = 6 and c = -3k

Now,

Sum of zeroes = (α+β) = -b/a = -6/k

Product of zeroes = αβ = c/a = -3k/k = -3

It is given that, the sum of zeroes is twice their product.

==> -6k = 2(-3)

==> -6k = -6

==> k = -6/-6

==> k = 1

Therefore, the value of k is 1.