The quadratic equation x^2-2kx+(k-1)=0 has roots α and β such that α^2+β^2=4.

Without solving the equation, find the possible values of the real number k

Solution:-

let us say p, q are the roots of the quadratic equation.

x^2-2kx+(k-1)=0

on comparison with the standard formula a=1 , b= -2k , c= k-1

we can check the discriminant b^2- 4ac to check the possible real values of “k”.

this article speaks more about the discriminant and its conditions

https://www.conceptmathhelp.com/finding-roots-quadratic-equation-using-quadratic-formula/

There would be real roots, the discriminant b^2- 4ac >= 0

let us check the same by substituting a,b,c values into the discriminant

b^2- 4ac =(-2k)^2 -4(1)(k-1)

=4k^2 -4k +4

=4(k^2-2k+1)

=4(k-1)^2

=4(k-1)^2

Since 4(k-1)^2 >= 0 always, we can conclude that there will be 2 real values of K as the solution.

Hope this helped. Please let me know for further questions.

let us say p, q are the roots of the quadratic equation.

x^2-2kx+(k-1)=0

on comparison with the standard formula a=1 , b= -2k , c= k-1

we can check the discriminant b^2- 4ac to check the possible real values of “k”.

this article speaks more about the discriminant and its conditions

https://www.conceptmathhelp.com/finding-roots-quadratic-equation-using-quadratic-formula/

There would be real roots, the discriminant b^2- 4ac >= 0

let us check the same by substituting a,b,c values into the discriminant

b^2- 4ac =(-2k)^2 -4(1)(k-1)

=4k^2 -4k +4

=4(k^2-2k+1)

=4(k-1)^2

=4(k-1)^2

Since 4(k-1)^2 >= 0 always ,we can conclude that there will be 2 real values of K as the solution.

Hope this helped. Please let me know for further questions.