Question

3x² - KX + 5 Find K if
equatoin has
Equal roots​

Answer 1

for equal root D=0

b^2-4ac=0

k^2-60=0

k= rt.60

Hope it helps

Answer 2

Given :

Find the remainder when P(y) = y2 – 3y -5 is divided by 2y-1.

The equation 3x^2 - kx + 5 has equal roots .

To find :

The value of k.

Solution :

If the quadratic equation has its roots real and equal it implies that

b^2 - 4ac = 0

Here ,

a = 3

b = -k

c = 5

${b}^{2} - 4ac = 0 \\ { - k}^{2} - 4 \times 3 \times 5 = 0 \\ {k}^{2} - 60 = 0 \\ {k}^{2} = 60 \\ k = \sqrt{60}$