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 $x=(3-sqrt(13))/2$ .

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