Question

How do you solve $3{{x}^{3}}-48x=0$?

Answer 1

Hint:We first take the common constants and variables from the equation $3{{x}^{3}}-48x=0$. 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 $3{{x}^{3}}-48x=0$.

Complete step by step answer:
We need to find the solution of the given equation $3{{x}^{3}}-48x=0$.
We will take the common constants and variables from the equation. It will be $3x$.
Now we have a quadratic equation $3x\left( {{x}^{2}}-16 \right)=0$.
Now we find the factorisation of the equation $\left( {{x}^{2}}-16 \right)=0$ using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Therefore, we get $\left( {{x}^{2}}-16 \right)={{x}^{2}}-{{4}^{2}}=\left( x+4 \right)\left( x-4 \right)$.
So, $3{{x}^{3}}-48x=3x\left( x+4 \right)\left( x-4 \right)$. Equating with 0 we get
We get the values of x as either $\left( x+4 \right)=0$ or $\left( x-4 \right)=0$. The value of x also can be 0.
This gives $x=0,-4,4$.

The given quadratic equation has 3 solutions and they are $x=0,\pm 4$.

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 $\left( {{x}^{2}}-16 \right)=0$.
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 $\left( {{x}^{2}}-16 \right)=0$. The values of a, b, c is $1,0,-16$ respectively.
We put the values and get x as $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times \left( -16 \right)}}{2\times 1}=\dfrac{\pm \sqrt{64}}{2}=\dfrac{\pm 8}{2}=\pm 4$.