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

 Jun 12, 2023
 #1
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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.

 Jun 12, 2023
 #2
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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.  

.

 Jun 12, 2023

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