Question

how to solve this quadratic eqution?find the range of values of "k" for which the roots of the equation x^2 + (2k+1)x+k^2 -3 = 0 , are not real. State the range of values of "k" for which the roots are real.

Answer 1

Use the quadratic formula or complete the square. Then look at the term inside the radical. The radical of a real number is only real, if that number is not negative! So solve the inequality >= 0 for the term inside the radical.

Answer 2

what is redicle?

Answer 3

The square root!

Answer 4

ok

Answer 5

The solutions of your quadratic equation are \[ -k-\frac{1}{2}\pm\frac{1}{2}\cdot\sqrt{4k+13} \].

Answer 6

oh thankx i am tring to solve it :)

Answer 7

k^2+(2 k+1) x+x^2-3 = 0
Expand out terms of the left hand side:
k^2+2 k x+x^2+x-3 = 0
Subtract x^2+x-3 from both sides:
k^2+2 k x = -x^2-x+3
Add x^2 to both sides:
k^2+2 k x+x^2 = 3-x
Factor the left hand side:
(k+x)^2 = 3-x
Take the square root of both sides:
abs(k+x) = sqrt(3-x)
Eliminate the absolute value:
k+x = -sqrt(3-x) or k+x = sqrt(3-x)
Subtract x from both sides:
k = -sqrt(3-x)-x or k+x = sqrt(3-x)
Subtract x from both sides:
k = -sqrt(3-x)-x or k = sqrt(3-x)-x

Answer 8

k=-+[sqrt(3-x)]-x
k=-+[sqrt(3-x)] -x, for 0<x<3, k=Real and for x>3,k=imaginary ans