# sum and product of roots

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.

1 Answers

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.