Question

How do you solve $2{{x}^{2}}-18=0$?

Answer 1

Hint: We first divide both sides of the equation by 2. Then we form the equation according to the identity ${{a}^{2}}-{{b}^{2}}$ to form the factorisation of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We place values $a=x;b=3$. The multiplied polynomials give value 0 individually. From that we find the value of x to find the solution of $2{{x}^{2}}-18=0$.

Complete step by step answer:
We need to find the solution of the given equation $2{{x}^{2}}-18=0$.
First, we divide both sides of the equation by 2 and get $\dfrac{2{{x}^{2}}-18}{2}=0\Rightarrow {{x}^{2}}-9=0$.
Now we have a quadratic equation ${{x}^{2}}-9=0$ which gives ${{x}^{2}}-{{3}^{2}}=0$.
Now we find the factorisation of the equation ${{x}^{2}}-{{3}^{2}}=0$ using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Therefore, we get
$\begin{align}
  & {{x}^{2}}-{{3}^{2}}=0 \\
 & \Rightarrow \left( x+3 \right)\left( x-3 \right)=0 \\
\end{align}$
We get the values of x as either $\left( x+3 \right)=0$ or $\left( x-3 \right)=0$.
This gives $x=-3,3$.
The given quadratic equation has 2 solutions and they are $x=-3,3$.

Note:
The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation $2{{x}^{2}}-18=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $2{{x}^{2}}-18=0$. The values of a, b, c are $2,0,-18$ respectively.
We put the values and get x as $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 2\times \left( -18 \right)}}{2\times 2}=\dfrac{\pm \sqrt{144}}{4}=\dfrac{\pm 12}{4}=\pm 3$.