For what value of k does the equation 2x^2 − 6x + k = 0 have exactly one solution?
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
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x = (6 / 4) = 3/2
Therefore, the value of k that makes the equation have exactly one solution is k = 3/2.
+1277
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 b2 – 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.
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