Question

How do you solve $3{y^2} - 12 = 0$?

Answer 1

Hint: In this question we have to solve the equation for $y$, the given equation is a quadratic equation as the degree of the highest exponent of $y$ is equal to 2. To solve the equation first take all $y$ terms to one side and all constants to the others side and further simplification of the equation $\begin{gathered}
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
    \\
\end{gathered} $ we will get the required $y$.

Complete step by step answer:
The given expression is a polynomial which are algebraic expressions that are composed of two or more algebraic terms, the algebraic terms are constant, variables and exponents, there are different types of polynomials which are differentiated with the help of degree of the polynomial. The polynomial of degree 2 is known as quadratic polynomial.
Given algebraic expression is $3{y^2} - 12 = 0$, and we have to solve the equation for $y$
The given polynomial is a quadratic polynomial as it has degree 2 in terms of $y$.
First add 12 to both sides of the equation, we get,
$ \Rightarrow 3{y^2} - 12 + 12 = 0 + 12$,
Now simplifying by eliminating the like terms, we get,
$ \Rightarrow 3{y^2} = 12$,
Now divide both sides of the equation with 3, we get,
$ \Rightarrow \dfrac{{3{y^2}}}{3} = \dfrac{{12}}{3}$,
Now simplifying we get,
$ \Rightarrow {y^2} = 4$,
Now applying square root to the both sides of the equation, we get,
$ \Rightarrow \sqrt {{y^2}} = \sqrt 4 $
Now simplifying we get,
$ \Rightarrow y = \pm 2$,

So, we get two values for $y$ and they are $ \pm 2$.

Note: We will solve the above equation using Quadratic formula, which is given by, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
Now here given equation is $3{y^2} - 12 = 0$,
Rewrite the equation as , $3{y^2} + 0y - 12 = 0$,
On comparing the equation with the basic quadratic equation i.e., $a{x^2} + bx + c = 0$,so here $x = y$,$a = 3$, $b = 0$ and $c = - 12$,
By substituting the values in the formula we get,
$ \Rightarrow y = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4\left( 3 \right)\left( { - 12} \right)} }}{{2\left( 3 \right)}}$,
Now simplifying we get,
$ \Rightarrow y = \dfrac{{ \pm \sqrt {144} }}{6}$,
Now further simplifying we get,
$ \Rightarrow y = \dfrac{{ \pm 12}}{6} = \pm 2$,
So we got the same values for $y$ in both the cases.