Question

Solve the following equation and find the values of m:
$\left( 2{{m}^{3}}+3m \right)\left( 5m-1 \right)=0.$

Answer 1

Hint: To solve the question given above, first we will find the total number of values of m we will get. The total number of values of m will be equal to greater power of m in the equation. Then, we will find all the values of m by factoring the above equation and then equating each factor term to zero.

Complete step-by-step answer:
Before we find the values of m or roots of the equation given above, we will first find out how many roots we will get. The total number of roots will be equal to the highest power of m in the equation $\left( 2{{m}^{3}}+3m \right)\left( 5m-1 \right)=0.$ For this, we will multiply both the terms on the left hand side. Thus we will get:
$\begin{aligned} & \left( 2{{m}^{3}}+3m \right)\left( 5m-1 \right)=0 \\\ & \Rightarrow 2{{m}^{3}}\left( 5m-1 \right)+3m\left( 5m-1 \right)=0 \\\ & \Rightarrow 10{{m}^{4}}-2{{m}^{3}}+15{{m}^{2}}-3m=0.........\left( 1 \right) \\\ \end{aligned}$
Here, we can see that the greater power of m is 4 so we will get roots ${{m}_{1}},{{m}_{2}},{{m}_{3}}\text{ }and\text{ }{{m}_{4.}}$ Now, the equation given in question is:
$\left( 2{{m}^{3}}+3m \right)\left( 5m-1 \right)=0.$
Now, we will take m common from the term $\left( 2{{m}^{3}}+3m \right).$Thus we will get: $\left( m \right)\left( 2{{m}^{2}}+3 \right)\left( 5m-1 \right)=0.$
In the above equation, we have the term $\left( 2{{m}^{3}}+3m \right)$ which do not have real factors. The imaginary factors of equation ${{a}^{2}}{{x}^{2}}+{{b}^{2}}$ are $\left( x+\dfrac{i6}{a} \right)\left( x-\dfrac{i6}{a} \right).$ So, after applying these form, we will get: $m\left( m+\dfrac{i\sqrt{3}}{\sqrt{2}} \right)\left( m-\dfrac{i\sqrt{3}}{\sqrt{2}} \right)\left( 5m-1 \right)=0$
Now, to find the values of m, we will equate each factor to zero separately. Thus ${{m}_{1}}$ can be obtained as follows: $\begin{aligned} & m=0 \\\ & \Rightarrow {{m}_{1}}=0 \\\ \end{aligned}$
Now, to find the value of ${{m}_{2}}$, we will equate the form $\left( m+\dfrac{i\sqrt{3}}{\sqrt{2}} \right)$ to zero. Thus, we get: $\begin{aligned} & m+\dfrac{i\sqrt{3}}{\sqrt{2}}=0 \\\ & m=-i\sqrt{\dfrac{3}{2}} \\\ & \Rightarrow m=-\sqrt{\dfrac{3}{2}}i \\\ \end{aligned}$
To find the value of ${{m}_{3}}$, we will equate the term $\left( m-\dfrac{i\sqrt{3}}{\sqrt{2}} \right)$ to zero. Thus, we get:
$\begin{aligned} & m+\dfrac{i\sqrt{3}}{\sqrt{2}}=0 \\\ & \Rightarrow m=i\sqrt{\dfrac{3}{2}} \\\ & \Rightarrow {{m}_{3}}=\sqrt{\dfrac{3}{2}i} \\\ \end{aligned}$
To find ${{m}_{4}}$, we will do:
$\begin{aligned} & 5m-1=0 \\\ & m=\dfrac{1}{5}\Rightarrow {{m}_{4}}=\dfrac{1}{5} \\\ \end{aligned}$
Thus the roots of equation $\left( 2{{m}^{3}}+3m \right)\left( 5m-1 \right)=0.$ are
$\begin{aligned} & {{m}_{1}}=0 \\\ & {{m}_{2}}=-\sqrt{\dfrac{3}{2}i} \\\ & {{m}_{3}}=\sqrt{\dfrac{3}{2}i} \\\ & {{m}_{4}}=\dfrac{1}{5} \\\ \end{aligned}$

Note: The imaginary roots of the above equation can also be found out with the help of quadratic formula. The quadratic formula is as shown: $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where x is the root of $a{{x}^{2}}+bx+c=0.$ In our case, the factor which will give imaginary roots is$\left( 2{{m}^{2}}+3 \right).$ Thus a=2, b=0 and c=3.
$\begin{aligned} & \Rightarrow m=\dfrac{-\left( 0 \right)\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow m=\dfrac{\pm \sqrt{-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow m=\pm \dfrac{2\sqrt{-\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow m=\pm \dfrac{\sqrt{\left( 2 \right)\left( 3 \right)i}}{\left( 2 \right)} \\\ & \Rightarrow m=\pm \left( \sqrt{\dfrac{3}{2}} \right)i \\\ & \Rightarrow m=\sqrt{\dfrac{3}{2}}i\text{ and m=-}\sqrt{\dfrac{3}{2}}i. \\\ \end{aligned}$