How do you solve #-x^2-12=-87#?

2 Answers
Mar 7, 2018

Answer:

#x=+-5sqrt(3)#

Explanation:

If
#color(white)("XXX")-x^2-12=-87#
then
#color(white)("XXX")x^2+12=87#

#color(white)("XXX")x^2=75#

#color(white)("XXX")x=+-5sqrt(3)#

Mar 7, 2018

Answer:

#x=+-5sqrt(3)#

Explanation:

Multiply all of both sides by (-1)

#+x^2+12=+87#

Subtract 12 from both sides

#x^2=87-12 = 75#

#x=sqrt(75)#

But 75 #->5xx15->5xx5xx3 = 5^2xx3#

#x=sqrt(5^2xx3)#

#x=5sqrt(3)# as an EXACT principle root answer (only positive)

However. This is a quadratic so it will have both positive and negative roots. The plot will cross the x-axis in 2 places

#x=+-5sqrt(3)#