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

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Answer 1 · 2017-08-08T07:53:00

The solutions of this equation are imaginary.

The roots of this equation are:

$x=-1.07i, x=0.548i$

The zeroes (solutions) occur when $y=0$ , so we have $−9/2x^2+3/2x–3/2=0$ .

Standard form for a quadratic equation is $ax^2+bx+c=0$ , so in this case, $a=-9/2$ , $b=3/2$ and $c=-3/2$ .

The quadratic formula is $x=(-b+-sqrt(b^2-4ac))/(2a)$

Substituting in the values of $a, b$ and $c$ , we have:

$x=(-3/2+-sqrt((3/2)^2-4xx(-9/2)xx(-3/2)))/(2(3/2))$
$=(-3/2+-sqrt((9/4-27)))/6$

Since the total under the square root sign yields a negative number, there are no real roots of the equation, only imaginary roots.

$sqrt(9/4-27)=4.97i$ where $i=sqrt(-1)$ .

That means the roots of this equation are:

$x=-1.07i, x=0.548i$

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