Quadratic Equations Question 1

What are the solutions to the equation $3x^2-36=0 $ ?

$ -2\sqrt{3} \text{ and } 2\sqrt{3} $

$ -12 \text { and } 12 $

$ -6 \text{ and } 6 $

$ -2 \text{ and } 2 $

Approach 1

$$3x^2-36=0$$ $$ 3x^2=36 $$ $$ x^2=12 $$ $$ x= \pm \sqrt{12} $$ $$ x= \pm 2\sqrt{3} $$ $$ x = \boxed{-2\sqrt{3} \text{ and } 2\sqrt{3}} $$

Approach 2

Although a bit tedious, we can substitute the values in the answer choices into the given equation. The first one works, so we can stop there:

$$ 3x^2-36=0 $$ $$ 3(\pm 2\sqrt{3})^2-36=0$$ $$ 3(4\cdot 3)-36=0$$ $$ 3(12)-36=0$$ $$ 36-36=0 \checkmark$$