Find k such that $3 x^{2} + 2 k x + x - k - 5$ has sum of zeros equal to half of the product.

Open in App

Solution

$3 x^{2}$ + (2k+1)x - (k+5)
sum of zeros = a+b = - $\frac{b}{a}$ = - $\frac{(2 k + 1)}{3}$
Product of zeros = ab = $\frac{c}{a}$ = $- \frac{(k + 5)}{3}$
sum of zeros = $\frac{1}{2} \times p r o d u c t$
$- \frac{(2 k + 1)}{3} = - \frac{1}{2} \times \frac{(k + 5)}{6}$
24k+12 = 3k+15
21k = 3
k = $\frac{1}{7}$

Suggest Corrections

Similar questions

k

3 x^{2} + 2 k x + x - k - 5

3 x^{2} + 2 k x + x - k - 5

The value of k such that $3 x^{2} + 2 k x + x - k - 5$ have the sum of the zeroes as half of their product,

Find the value(s) of k so that the quadratic equation $3 x^{2} - 2 k x + 12 = 0$ has equal roots.

If the sum of the zeros of a polynomial $x^{2} - (k + 3) x + (5 k - 3)$ is equal to one fourth of the product of zeros find the value of $k$ .

Join BYJU'S Learning Program