For what value of k does the equation 2x^2 − 6x + k = 0 have exactly one solution?

Guest Jun 12, 2023

#1**0 **

For a quadratic equation of the form ax^2 + bx + c = 0, the number of solutions is given by the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

The equation has exactly one solution when the discriminant b^2 - 4ac = 0. In this case, the quadratic formula reduces to

x = (-b / 2a)

Substituting the values from the given equation, we get

Code snippet

x = (6 / 4) = 3/2

Therefore, the value of k that makes the equation have exactly one solution is k = 3/2.

Guest Jun 12, 2023

#2**+1 **

In regard to #1 above "...the value of k that makes the equation have exactly one solution is k = 3/2."

I think that solution was for the value of x, instead of for the value of k.

I followed as far as "...one solution when the discriminant b^2 - 4ac = 0." So far, so good.

At that point, I think I would substitute values in b^{2} – 4ac = 0

(6)^{2} – (4)(2)(k) = 0

36 – 8k = 0

8k = 36

k = 36/8

**k = 9/2**

I don't know how to check the veracity

of this answer. If I'm wrong, I would truly

appreciate someone showing me where.

_{.}

Bosco Jun 12, 2023