Question

Solve the quadratic equation $3{{x}^{2}}-23x+40=0$ by method of factorization.

Answer 1

Hint: Now we are given with a quadratic equation in x. To find the solution of the equation we will solve the equation by factorization method. Now we will split the middle terms such that the product of the two terms is the product of the first term and last term. Hence we will simplify the equation further by taking terms common. Hence on further simplification we will find the solution of the equation.

Complete step by step solution:
Now let us first consider the quadratic equation $3{{x}^{2}}-23x+40=0$ .
Now to solve the equation we will use the factorization method. Hence we will first split the middle term in two parts such that the product of the two terms is the same as the product of the first term and the last term. Now here the middle term is $-23x$ and the first term and last term are $3{{x}^{2}}$ and 40 respectively.
Now we can split the term $-23x$ as $-15x-8x$ since $\left( -15x \right)\left( -8x \right)=120{{x}^{2}}=\left( 3{{x}^{2}} \right)\left( 40 \right)$
Now rewriting the equation we get,
$\Rightarrow 3{{x}^{2}}-15x-8x+40=0$
Now let us take 3x common from the first two terms and -8 common from the last two terms.
$\Rightarrow 3x\left( x-5 \right)-8\left( x-5 \right)=0$
Now again taking $\left( x-5 \right)$ common we get,
$\Rightarrow \left( 3x-8 \right)\left( x-5 \right)=0$
Now if multiplication of two terms is 0 then one of the terms must be 0.
Hence either $\left( 3x-8 \right)=0$ or $x-5=0$ .
Hence we get x = 5 or $x=\dfrac{8}{3}$

Hence the roots of the equation are $5$ and $\dfrac{8}{3}$

Note: Now note that we can also find the solution of the equation by using the formula of roots for quadratic equation. We have the roots of the quadratic equation $a{{x}^{2}}+bx+c$ is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence substituting the values of a, b and c we get the solution of the equation.