Question

Prove that the roots of x^2+(1-k)x+k-3=0 are real for all real values of k

Answer 1

Answer:

i think its intermediate question right.....

Answer 2

For the roots of the quadratic equation x²+(1-k)x+k-3=0 to be real the value of the determinant should be always greater than equal to zero.

Determinant = D = (1-k)² - 4×1×(k-3)

=> (1-k)² - 4×1×(k-3)

=> 1+k²-2k-4k+12

=> k² -6k +13 >0

Value of D = k² -6k +13 is always greater than zero because the roots of equation k² -6k +13 (D<0) are imaginary as it's determinant is negative.

As D>0 for the equation x²+(1-k)x+k-3=0 , the roots of the equation are real for all real values of k.