49.7k views
2 votes
given that the quadratic equation x^2-x+k=(k+1)x-5, has two equal roots, state the possible value of k.

1 Answer

3 votes

Answer:If the quadratic equation x^2 - x + k = (k + 1)x - 5 has two equal roots, then the discriminant of the quadratic equation is zero.

The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac.

In this case, the quadratic equation is x^2 - x + k = (k + 1)x - 5, which can be rearranged as x^2 - (k + 2)x + (k - 5) = 0.

Therefore, the discriminant is:

(k + 2)^2 - 4(1)(k - 5) = 0

Simplifying and solving for k, we get:

k^2 + 4k + 4 - 4k + 20 = 0

k^2 + 24 = 0

(k + 2√6i)(k - 2√6i) = 0

where i is the imaginary unit.

Therefore, the possible values of k are -2√6i and 2√6i.

However, since the problem statement does not mention imaginary roots, the possible value of k that satisfies the condition of having two equal roots is k = -2.

If we substitute k = -2 into the quadratic equation, we get:

x^2 - x - 2 = -x - 5

x^2 = 4

x = ±2

Therefore, the two equal roots of the quadratic equation are x = 2 and x = -2, and k = -2 satisfies the condition of having two equal roots.

Explanation:

User Robin Kanters
by
8.4k points