Justify if completing the square is a good method for solving when the Discriminant is negative. Use any of your three functions as an example and respond in complete sentences.

Function h(x) = 2x^2 + 1x + 3

Question

Justify if completing the square is a good method for solving when the Discriminant is negative. Use any of your three functions as an example and respond in complete sentences.

Function h(x) = 2x^2 + 1x + 3

anonymous · Answer

Last question for today and I'm stuck on it lol

mathstudent55 · Answer

First you need to set the function equal to zero.

$h(x) = 2x^2 + 1x + 3$

$ 2x^2 + 1x + 3 =0 $

mathstudent55 · Answer

To complete the square, the leading coefficient (the coefficient of $x^2$) must be 1.
To do that, we divide both sides by 2.

$x^2 + \dfrac{1}{2}x + \dfrac{3}{2} =0 $

mathstudent55 · Answer

Are you following?

anonymous · Answer

yeah

mathstudent55 · Answer

The next step is to move the constant term (the term with no x) to the right side, and leave a space for the completing of the square term we are going to add to both sides.

mathstudent55 · Answer

$x^2 + \dfrac{1}{2} x ~~~~~~~~ = -\dfrac{3}{2} $

mathstudent55 · Answer

Now we are actually going to add the term that completes the square.

That term is found by taking the x-term's coefficient, in this case $\dfrac{1}{2}$, dividing it by 2 getting $\dfrac{1}{4}$, and then squaring it to get $ \dfrac{1}{16} $.

mathstudent55 · Answer

Now we add the $ \dfrac{1}{16} $ to both sides.

$x^2 + \dfrac{1}{2} x + \dfrac{1}{16} = -\dfrac{3}{2} + \dfrac{1}{16}$

mathstudent55 · Answer

Now I need to ask you a question. Have you learned imaginary numbers yet? If the discriminant of a quadratic equation is negative, which means it doesn't have real roots, have you learned how to deal with imaginary or complex roots? Or do you just state no real roots, and you don't do anything else?

anonymous · Answer

I have learned a little bit about imaginary numbers but I'm not very good with them

mathstudent55 · Answer

Ok, let's go one with the problem, then.

anonymous · Answer

Yeah I need to continue

anonymous · Answer

Man you got one heck of an answer going lol

mathstudent55 · Answer

I noticed a mistake above.
I've erased a few responses above. The last equation I left above is correct. I will continue from there.

Since we completed the square on the left side, the left side is just the square of a binomial. On the right side, we just need to add the fractions. We need a common denominator. The LCD is 16.

mathstudent55 · Answer

$x^2 + \dfrac{1}{2} x + \dfrac{1}{16} = -\dfrac{3}{2} \cdot \dfrac{8}{8}+ \dfrac{1}{16}$

$x^2 + \dfrac{1}{2} x + \dfrac{1}{16} = -\dfrac{24}{16} + \dfrac{1}{16}$

$x^2 + \dfrac{1}{2} x + \dfrac{1}{16} = -\dfrac{23}{16} $

$ \left( x + \dfrac{1}{4} \right)^2 =  -\dfrac{23}{16} $

$ x + \dfrac{1}{4} =  \pm \sqrt{-\dfrac{23}{16}} $

$ x + \dfrac{1}{4} =  \pm i{\dfrac{\sqrt{23}}{\sqrt{16}}} $

$ x + \dfrac{1}{4} =  \pm {\dfrac{i\sqrt{23}}{4}} $

$ x = -\dfrac{1}{4}  \pm {\dfrac{i\sqrt{23}}{4}} $

That is the solution.

As you can see, it can be done using the completing the square method, but I think it's quicker using the quadratic formula.

anonymous · Answer

Thanks man!  Now I just have to write all this down lol

mathstudent55 · Answer

You're welcome.
Go over the steps and explanations and ask questions if you have any.