Question

How do you solve $9{{x}^{2}}=27x$?

Answer 1

Hint: We first keep the variables in one side and factorise the quadratic equation. We take common terms out to form the multiplied forms. We get multiplication of two terms which gives 0. The individual terms will be 0 and from those polynomials we get the roots.

Complete step by step answer:
We need to find the solution of the given equation $9{{x}^{2}}=27x$.
First, we divide both sides of the equation by 9 and get $\dfrac{9{{x}^{2}}}{9}=\dfrac{27x}{9}\Rightarrow {{x}^{2}}=3x$.
Now we keep the variables in one side and get ${{x}^{2}}-3x=0$
Now we have a quadratic equation ${{x}^{2}}-3x=0$.
We try to take the common numbers out.
For ${{x}^{2}}-3x$, we take $x$ and get $x\left( x-3 \right)$.
Therefore, $x\left( x-3 \right)=0$. Multiplication of two terms gives 0. This means at least one individual term to be 0.
We get the values of x as either $x=0$ or $\left( x-3 \right)=0$.
This gives $x=0,3$.

The given quadratic equation $9{{x}^{2}}=27x$ has 2 solutions and they are $x=0,3$.

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 $9{{x}^{2}}=27x$.
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 $9{{x}^{2}}-27x=0$. The values of a, b, c is $9,-27,0$ respectively.
We put the values and get $x=\dfrac{-\left( -27 \right)\pm \sqrt{{{\left( -27 \right)}^{2}}-4\times 9\times 0}}{2\times 9}=\dfrac{27\pm \sqrt{{{27}^{2}}}}{18}=\dfrac{27\pm 27}{18}=0,3$.