Question

if roots of the quadratic equation x^2-kx+289=0 are real and equal then value of k equals​

Answer 1

Hey ! Mate!

Given equation is: kx(x−2)+6=0

kx2−2kx+6=0

Then a=k,b=−2k,c=6

Then b2−4ac=0

So (−2k)2−4k(6)=0

4k2−24k=0

4k(k−6)=0

k=0 and k=6

Put value of k

Then 6x2−12x+6=0

6x2−6x−6x+6=0

6x(x−1)−6(x−1)=0

∴x=1

Hope it Helps you !

Mark it as Brainliest Answer!

Answer 2

Step-by-step explanation:

a + a = k

2a = k

&. a*a = 289

a² = 289

a = √289 = 17

So , k = 2(17) = 34