Question

How do you solve the quadratic equation ${{x}^{2}}=32$?

Answer 1

We take all the variables and the constants all together. 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 $a=x;b=\sqrt{32}$. The multiplied polynomials give value 0 individually. From that we find the value of x to find the solution of ${{x}^{2}}=32$.

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

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 ${{x}^{2}}-32=0$.
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 ${{x}^{2}}-32=0$. The values of a, b, c is $1,0,-32$ respectively.
We put the values and get x as $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times \left( -32 \right)}}{2\times 1}=\dfrac{\pm \sqrt{128}}{2}=\dfrac{\pm 8\sqrt{2}}{2}=\pm 4\sqrt{2}$.