How do you solve #4x^2-6=74#?

Question

3197 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-03T21:11:59

#x=2sqrt5#

We can begin by adding #6# to both sides of this equation. Doing this, we get:

#4x^2=80#

We want to isolate #x#, so the next best step would be to divide both sides by #4#. We get:

#x^2=20#

Now, we can take the square root of both sides to get:

#x=+-sqrt20# (We have a positive and negative answer because squaring a negative number will also result in a positive number)

We can factor out a perfect square from #+-sqrt20#.
#20# is equal to #4*5#, so #sqrt20# will be equal to #sqrt4*sqrt5#. Now, we have:

#x=+-(color(blue)(sqrt4)*color(red)sqrt5)#

This simplifies to:

#x=+-color(blue)(2)color(red)sqrt5#