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#