Question

How do you solve ${{x}^{2}}-180=0$?

Answer 1

Hint:We first keep the variable and other constants on one side. We divide both sides of the equation by the constant of the coefficient of ${{x}^{2}}$. 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 $a=x;b=6\sqrt{5}$. The multiplied polynomials give value 0 individually. From that we find the value of $x$ to find the solution of ${{x}^{2}}-180=0$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}-180=0$.

The coefficient of the term ${{x}^{2}}$ is 1 and that’s why we don’t need to divide anything.

Now we have a quadratic equation ${{x}^{2}}-180=0$.

Now we find the factorisation of the equation ${{x}^{2}}-180=0$ using the identity of
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.

We know that 180 is the square of $6\sqrt{5}$ where $180={{\left( 6\sqrt{5} \right)}^{2}}$.
Therefore, we get
$\begin{align}
& {{x}^{2}}-180=0 \\
& \Rightarrow {{x}^{2}}-{{\left( 6\sqrt{5} \right)}^{2}}=0 \\
& \Rightarrow \left( x+6\sqrt{5} \right)\left( x-6\sqrt{5} \right)=0 \\
\end{align}$

We have multiplication of two polynomials which gives 0. This means at least one of them has to be
0. We get the values of $x$ as either $\left( x+6\sqrt{5} \right)=0$ or $\left( x-6\sqrt{5} \right)=0$.

This gives $x=-6\sqrt{5},6\sqrt{5}$.

The given quadratic equation has two solutions and they are $x=\pm 6\sqrt{5}$.

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 two roots. Cubic polynomials have three. It can be both real and imaginary roots.

We can also apply the quadratic equation formula to solve the equation ${{x}^{2}}-180=0$. 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 ${{x}^{2}}-180=0$. The values of a, b, c is $1,0,-180$ respectively.

We put the values and get x as $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times \left( -180
\right)}}{2\times 1}=\dfrac{\pm \sqrt{720}}{2}=\dfrac{\pm 12\sqrt{5}}{2}=\pm 6\sqrt{5}$.