# Prove that, for all real values of k, the roots of the quadratic equation x^2-(2+k)x-3=0 are real.?

show further that the roots are of opposite signs. Does anyone know how to do this???? I'd appreciate any help!!

### 2 Answers

- ?Lv 41 decade agoBest Answer
You approach a question like this by using the quadratic equation (for quadratic forms like ax^2+bx+c) and in particular the sqrt in the quadratic equation, which is called the discriminant:

discriminant: b^2-4ac

Now when the discriminant is positive there are 2 real solutions to the equation, when the discriminant is equal to zero there is 1 real solution, and when the discriminant is negative there are no solutions to the equation (this is because in the quadratic formula the discriminant is under a sqrt sign and you cannot take the sqrt of a negative number - thus no solution is possible).

2 solutions: discriminant>0

1 solution: discriminant=0

0 solutions: discriminant<0

In our question our a, b, & c are as follows:

a=1

b=-(2+k)

c=-3

discriminant:

[-(2+k)]^2 - 4(1)(-3) = (k+2)^2+12 > 0 (always positive as a square plus a positive number is positive)

after expanding:

=k^2+4k+16 which is always positive (this can be shown by graphing or by again computing the discriminant and showing this new quadratic has no solutions -> meaning it never crosses the x-axis and is always positive).

- mchaleLv 43 years ago
If the quadratic equation ax^2+bx+c = 0 has no genuine roots, then b^2 - 4ac < 0 With 7x^2 + kx + 8 = 0, b^2 - 4ac = ok^2 - 4(7)(8) = ok^2 - 224 for this reason for no genuine roots, ok^2 - 224 < 0 i.e. -4?14 < ok < 4?14 or ok lies interior the era (-4?14, 4?14)