How do you solve # x^2+18=-81#?

2 Answers
Apr 2, 2016

#x^2=-99=>x=+-sqrt(-99)=+-3sqrt(11)i#

Apr 2, 2016

Answer:

#color(blue)(=>x=+-3isqrt(11))#

Explanation:

Given:#" "color(brown)(x^2+18=-81)#

If so chosen you could write this as #y=x^2+99#

#color(blue)("Solving for "x" where the given equation is true")#

Subtract #color(blue)(18)# from both sides

#" "color(brown)(x^2+18color(blue)(-18)=-81color(blue)(-18))#

#" "x^2+0=-99#

know: #color(green)(3xx33=99)" so "color(blue)(3xx3xx11=99)" so "color(red)(3^2xx11=99)#

But we have negative 99 so we have #(-1)xx3^2xx11#

Thus our equation becomes

#x^2=(-1)xx3^2xx11#

Take the square root of each side

#x=+-sqrt((-1)xx3^2xx11)#

#=> x=+-3xxsqrt(11)sqrt(-1)#

But #sqrt(-1)=i#

#color(blue)(=>x=+-3isqrt(11))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that the graph does not cross the x-axis.

But by writing the equation as #x^2+18=-81#

#y=x^2+99#

So to solve for #x# we have to write

#y=0=x^2+99# but for #y=0# implies that the graph crosses the x-axis when in fact it does not. So to be able to solve for #y=# we introduce 'complex numbers'. Hence the #sqrt(-1)#

Tony B