How do you find the zeros of # y = 3/2x^2 – 9/2x - 3/2# using the quadratic formula?

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Answer 1 · 2017-12-13T18:39:36

#x=(3+sqrt(13))/2# or #x=(3-sqrt(13))/2#.

Start with #y=3/2x^2-9/2x-3/2#.

To find the zeros set #y# equal to 0:

#3/2x^2-9/2x-3/2=0#

Since the equation equals 0 we can multiply through by 2 to clear the denominators:

#3x^2-9x-3=0#

Now notice we can divide by 3 to further simplify:

#x^2-3x-1=0#

Now we have a quadratic equation with #a=1#, #b=-3#, and #c=-1# and the quadratic formula is #x=(-bpmsqrt(b^2-4ac))/(2a)#.

So we have:

#x=(-(-3)+sqrt((-3)^2-4(1)(-1)))/(2(1))=(3+sqrt(13))/2#

or we have:

#x=(-(-3)-sqrt((-3)^2-4(1)(-1)))/(2(1))=(3-sqrt(13))/2#

So, #x=(3+sqrt(13))/2# or #x=(3-sqrt(13))/2#.