How do you solve #4x²-12=0#?

2 Answers
Mar 20, 2018

Answer:

#x = +- 3#

Explanation:

#4x^2-12=0#

#=>4(x^2-3)=0#

#=>x^2-3 = 0#

#=>x^2 = 3#

#=>x = +- 3#

Answer:

#x = +-sqrt3#

Explanation:

Step 1:
Put all terms with #x# on one side of the equation and all terms without #x# on the other side of the equation.
#4x^2 - 12 = 0#
#4x^2 = 12#

Step 2:
Divide the equation (both sides) by largest common coefficient of 4.
#x^2 = 3#

Step 3:
Square the equation (both sides) to obtain #x# on one side as the subject of the equation (what you want to find).
Because 3 cannot be square-rooted to a whole answer, leave your answer in the root form.
#x = +-sqrt3#

Note:
In Step 2, you divided the equation by the coefficient 4. Take note that you cannot divide an equation by what you are finding (which in this case is #x#), especially if you have a quadratic equation (an equation with #x^n# where #n# is any real number).
In Step 3, remember to always add the #+-# after square-rooting unless you are doing an equation for real-world values (where there are no negative answers).