Question

Find the root of the quadratic equation $3{{\left( x-4 \right)}^{2}}-5(x-4)$ by factorisation.

Answer 1

Hint: Here, first consider the quadratic equation $3{{\left( x-4 \right)}^{2}}-5(x-4)=0$ and then take the common factor $x-4$ outside to make it in a simpler form. Now, we will get two factors and then by equating the factors to zero we will get the roots of the quadratic equation.

Complete step-by-step answer:
Here, we are given the quadratic equation $3{{\left( x-4 \right)}^{2}}-5(x-4)=0$
Now, we have to find the roots of the quadratic equation by factorisation.
We know that the polynomial equation with highest degree 2 is called the quadratic equation. It is expressed in the form of $a{{x}^{2}}+bx+c=0$, where a, b and c are the constant terms. Since, the quadratic equation is a second degree equation, it will have two roots.
Here, we have the equation $3{{\left( x-4 \right)}^{2}}-5(x-4)=0$.
Now, in the above equation $x-4$ is the common factor of both the terms. Hence, we can take it outside. Then the equation,
$\begin{align}
  & 3{{\left( x-4 \right)}^{2}}-5(x-4)=(x-4)\left( 3(x-4)-5 \right) \\
 & \Rightarrow 3{{\left( x-4 \right)}^{2}}-5(x-4)=(x-4)\left( 3\times x+3\times -4-5 \right) \\
 & \Rightarrow 3{{\left( x-4 \right)}^{2}}-5(x-4)=(x-4)\left( 3x-12-5 \right) \\
 & \Rightarrow 3{{\left( x-4 \right)}^{2}}-5(x-4)=(x-4)\left( 3x-17 \right) \\
\end{align}$
We have $3{{\left( x-4 \right)}^{2}}-5(x-4)=0$
Therefore, we can write in the form:
$(x-4)\left( 3x-17 \right)=0$
We know that if $ab=0$ then either a = 0, b = 0 or both zero.
Hence we can write:
$x-4=0$ or $3x-17=0$
For $x-4=0$, take -4 to the right side then -4 becomes 4,
$\Rightarrow x=4$
Now, consider the equation $3x-17=0$
Take -17 to the right side, then -17 becomes 17,
$\Rightarrow 3x=17$
Next by cross multiplication,
$\Rightarrow x=\dfrac{17}{3}$
Therefore, we can say that the two roots of the quadratic equation $3{{\left( x-4 \right)}^{2}}-5(x-4)=0$ are $x=4$ and $x=\dfrac{17}{3}$.

Note: There are many ways to find the roots of the quadratic equation. Consider the quadratic equation of the form $a{{x}^{2}}+bx+c=0$, then the sum of the roots is given as $\dfrac{-b}{a}$ and the product of the roots is given as $\dfrac{c}{a}$.Then we can find the two roots. Otherwise, we can directly apply the quadratic formula to obtain the roots and the formula is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$