Question

Find the value of x.
$\dfrac{1}{15}{{x}^{2}}+\dfrac{5}{3}=\dfrac{2}{3}x$

Answer 1

Hint: In such a type of question first reduce it to the simplest form. As the given equation has the degree 2 so reduce the above equation in $a{{x}^{2}}+bx+c=0$then find the solution of given quadratic equation by using the roots of equation
${{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step-by-step answer:
The given equation is
$\dfrac{1}{15}{{x}^{2}}+\dfrac{5}{3}=\dfrac{2}{3}x$
By taking LCM we can write it as
$\dfrac{{{x}^{2}}+\left( 5 \right)(5)}{15}=\dfrac{2}{3}x$
On cross multiplication we can write it as
$3({{x}^{2}}+25)=\left( 2 \right)(15)x$
Dividing both side by 3 we can write it as
\[\begin{align}
  & \dfrac{3({{x}^{2}}+25)}{3}=\dfrac{\left( 2 \right)(15)x}{3} \\
 & \Rightarrow {{x}^{2}}+25=10x \\
 & \Rightarrow {{x}^{2}}-10x+25=0 \\
\end{align}\]
Now we get the standard quadratic equation so on comparing we can write
$a=1,b=-10,c=25$
Hence, we can use the formula in order to find the roots of quadratic equation
${{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Putting the value of $a,b,c$in the above formula we have
\[\begin{align}
  & {{x}_{1,2}}=\dfrac{-(-10)\pm \sqrt{{{\left( -10 \right)}^{2}}-4(1)(25}}{2(1)} \\
 & \Rightarrow {{x}_{1,2}}=\dfrac{-(-10)\pm \sqrt{100-100}}{2} \\
 & \Rightarrow {{x}_{1,2}}=\dfrac{-(-10)}{2} \\
 & \Rightarrow {{x}_{1,2}}=5 \\
\end{align}\]
Hence the solution of the quadratic equation is $x=5$.

Note: if $a{{x}^{2}}+bx+c=0$is a quadratic equation then number of roots and its nature depends the value of D
$D={{b}^{2}}-4ac$
If $D>0,$We have two distinct roots
If $D=0,$We have one root, that is roots are coincident
If $D<0,$We have no real roots, both roots are imaginary.
In the above question D=0 so we have only one root.