Determine the value(s) of k for which
x^2+(k-2)x-2k=0
has equal and real roots.
a x^2 + bx + c = 0 has two different roots if the discriminant D defined as:
D = b^2 - 4 a c does not equal zero.
If D = 0 then there is one root. That root is then real if b/a is real. In this case this means that:
(k-2)^2 + 8k = 0 --->
(k+2)^2 = 0 --->
k = -2
k must also be real and -2 is real, so k = -2 is a valid solution.
its -3
To determine the value(s) of k for which the equation x^2 + (k-2)x - 2k = 0 has equal and real roots, we need to find the discriminant D first.
For a quadratic equation in the form ax^2 + bx + c = 0, the discriminant is given by D = b^2 - 4ac.
In this case, the quadratic equation is x^2 + (k-2)x - 2k = 0, so a = 1, b = (k-2), and c = -2k.
We substitute these values into the discriminant formula:
D = (k-2)^2 - 4(1)(-2k)
Expanding and simplifying this expression, we get:
D = k^2 - 4k + 4 + 8k
D = k^2 + 4k + 4
To have equal and real roots, the discriminant D must be equal to zero, as stated earlier.
Therefore, we set the discriminant equal to zero:
k^2 + 4k + 4 = 0
Now we solve this equation for k. Factoring this quadratic equation, we get:
(k + 2)^2 = 0
Taking the square root of both sides, we have:
k + 2 = 0
Solving for k, we find:
k = -2
So the value of k for which the equation has equal and real roots is k = -2.