Question

Values of K for which the quadratic equation \[2{{x}^{2}}-kx+k=0\] has equal roots is/are
A. 0
B. 4
C. 8
D.0,8

Answer 1

Hint: For any quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] its roots are calculated by the standard formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] and if we put this expression \[\sqrt{{{b}^{2}}-4ac}=0\] then we can see that now both roots are equal. And both root value equals to \[x=\dfrac{-b}{2a}\], SO on putting expression \[\sqrt{{{b}^{2}}-4ac}=0\]we got a relation in k and on solving we get the values of k.

Complete step-by-step answer:
We are given an quadratic expression as \[2{{x}^{2}}-kx+k=0\] and We have been asked to find the unknown value of k and condition for which is both the roots of this quadratic equation are equal .So for that First we will compare this given equation with general quadratic expression which is
\[a{{x}^{2}}+bx+c=0\], \[2{{x}^{2}}-kx+k=0\]
On comparing we get values of a, b, c as \[a=2,b=-k,c=k\]
And we know that condition for roots of quadratic to be equal is \[\sqrt{{{b}^{2}}-4ac}=0.....(1)\]
So, putting values of a, b, c in equation (1) we get expression as
\[\sqrt{{{(-k)}^{2}}-4(2)(k)}=0\] on solving it will look like \[\sqrt{{{k}^{2}}-8k}=0\]
Solving it further gives \[{{k}^{2}}-8k=0\] on factoring it become \[k(k-8)=0\],
So, we got the value k as 0 and 8 which satisfies it

Hence the values of K are 0,8 so answer is option (D)

So, the correct answer is “Option D”.

Note: If the condition is given that roots are real so the condition for that will be \[{{b}^{2}}-4ac\ge 0\]similarly for imaginary roots condition is \[{{b}^{2}}-4ac\le 0\].The actual meaning of having a single root or both roots common for a quadratic equation is that ,the graph of parabola will touch x at one point only and if it has real roots it will intersect at two points and similarly if imaginary roots then graph will not not x axis.