How do you find the zeros, real and imaginary, of #y=x^2-x+7# using the quadratic formula?

1 Answer
May 3, 2017

Answer:

Plug in the given values for #a#, #b#, and #c# into the formula, and solve.
#x=(1+-3isqrt(3))/(2)#.

Explanation:

Finding the zeroes (or roots) of a quadratic equation means solving for #x# when #y=0#. In other words, we want to know what values of #x# we would have to plug into #x^2-x+7# in order for it to simplify to #0.#

To do this, there are some shortcuts that might be easy to see, but the quadratic formula is always an option. This formula states that, if we want #ax^2+bx+c# to equal #0,# then the #x#-values that make this true will be

#x=(–b+-sqrt(b^2-4ac))/(2a)#

So, for any specific quadratic expression (like #x^2-x+7#), all we need to do is plug the given values for #a, b,# and #c# into the quadratic formula.

For #x^2-x+7#, we have

#a=1# (from the #1x^2#),
#b=–1# (from the #-1x#), and
#c=7# (from the #+7#).

Plugging these into the quadratic formula gives

#x=(–color(purple)b+-sqrt(color(blue)b^2-4color(orange)acolor(green)c))/(2color(orange)a)#

#x=(–color(blue)((–1))+-sqrt(color(blue)((–1))^2-4color(orange)((1))color(green)((7))))/(2color(orange)((1)))#

#x=(1+-sqrt(1-28))/(2)#

#x=(1+-sqrt(–27))/(2)#

#x=(1+-3sqrt(–3))/(2)#

Because we get a negative under the square root, the zeroes are imaginary. Recalling that #i^2=–1#, we get

#x=(1+-3isqrt(3))/(2)#.