If the roots of the quadratic equation $$x^2-kx+169=0$$​ are equal, find the value of $$k$$​.

Correct option is A

Given:

Given quadratic equation =$$x^2-kx+169=0$$​​

Formula Used:

For finding the nature of roots we use determinant D = $$b^2 - 4ac$$​ for any equation in form of $$ax^2 +bx +c = 0$$​​

If D =0 then real and equal roots

If D<0 then roots are imaginary or no real roots

If d>0 then roots are real and distinct

Solution:

Here in the equation $$x^2-kx+169=0$$​​

Here a =1 , b=-k and c = 169

So D =$$(-k)^2 - 4(1)(169) = k^2 - 676$$​

​$$k^2 = 676$$​​

​$$k= \sqrt{676}$$​​

k = $$\pm 26$$​​

Hence option (a) is correct