Question

Solve the given quadratic equation $2{{x}^{2}}=32$?

Answer 1

We first keep the variable on one side and all the other variables on the other side. We 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 $a=x;b=4$. The multiplied polynomials give value 0 individually. From that we find the value of x to find the solution of $2{{x}^{2}}=32$.

Complete step-by-step solution:
We need to find the solution of the given equation $2{{x}^{2}}=32$.
First, we divide both sides of the equation by 2 and get ${{x}^{2}}=\dfrac{32}{2}=16$.
Now we have a quadratic equation ${{x}^{2}}=16$ which gives ${{x}^{2}}-16=0$.
Now we find the factorisation of the equation ${{x}^{2}}-{{4}^{2}}=0$ using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Therefore, we get
$\begin{aligned} & {{x}^{2}}-{{4}^{2}}=0 \\\ & \Rightarrow \left( x+4 \right)\left( x-4 \right)=0 \\\ \end{aligned}$
We get the values of x as either $\left( x+4 \right)=0$ or $\left( x-4 \right)=0$.
This gives $x=-4,4$.
The given quadratic equation has 2 solutions and they are $x=-4,4$.

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 $2{{x}^{2}}=32$.
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 $2{{x}^{2}}-32=0$. The values of a, b, c are $2,0,-32$ respectively.
We put the values and get x as $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 2\times \left( -32 \right)}}{2\times 2}=\dfrac{\pm \sqrt{256}}{4}=\dfrac{\pm 16}{4}=\pm 4$.