Question

If the
sum
Zeros
of
the
quadratic polynomial kx² + 2x +3k
is
equal to
to their product find
the value
of k​

Answer 1

GIVEN :–

• A quadratic equation kx² + 2x +3k = 0 have two roots α and β .

• Sum of roots (α + β) = Product of roots (αβ) .

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

▪︎ We know that –

=> Sum of roots = α + β = -( coffieciant of x)/(coffieciant of x²)

=> α + β = - (2/k) ——————eq.(1)

▪︎ And –

=> Product of roots = αβ = (constant term)/(coffieciant of x²)

=> αβ = (3k)/(k)

=> αβ = 3 ——————eq.(2)

▪︎ Now According to the eq.(1) & eq.(2) –

=> -(2/k) = 3

=> -2 = 3k

=> k = - (2/3)

● Hence , The value of k is -(⅔) .

Answer 2

Given :—

Sum of zeros of polynomial = product of zeros of polynomial.

$\sf \therefore \alpha + \beta = \alpha \beta$

We know that:—

Sum of zeros of polynomial = -b/a

Sum of zeros of polynomial = -b/aProduct of zeros of polynomial = c/a

To find :—

Value of k.

Solution :—

$\sf here \: a = k, \: b = 2 \: and \: c = 3k$

$\sf \frac{ - b}{a} = \frac{c}{a}$

$\sf - b = c$

$\sf - (2) = 3k$

$\sf 3k = - 2$

$</strong><strong>\sf </strong><strong>\</strong><strong>t</strong><strong>h</strong><strong>e</strong><strong>r</strong><strong>e</strong><strong>f</strong><strong>o</strong><strong>r</strong><strong>e</strong><strong> </strong><strong>k = \frac{ - 2}{3}$