How do you find the zeros of # y = x^2 - 5x -13 # using the quadratic formula?

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Answer 1 · 2018-03-13T17:18:13

The quadratic formula always gives two answers. In this case, #x=6.8875# and #x=-1.8875#

To find the 'zeros', you want to solve the function for when y=0:

#0=x^2-5x-13#

There is a general form for this, which is what we will be using to describe the quadratic formula:

#0=ax^2+bx+c#

The quadratic formula looks like this:

#(-b+-sqrt(b^2-4ac))/(2a)#

For this case:
#a=1#
#b=-5#
#c=-13#

Now, let's plug those into the formula:

#x=(-(-5)+-sqrt((-5)^2-4(1)(-13)))/(2(1))#

#x=(5+-sqrt(25-4*(-13)))/2#

#x=(5+-sqrt(25+52))/2 rArr x=(5+-sqrt(77))/2 #

Now, we solve for both scenarios of x:

#color(red)(x=(5+sqrt(77))/2 rArr x=6.8875#

#color(blue)(x=(5-sqrt(77))/2 rArr x=-1.8875#