Question

How do you solve \[-10{{x}^{2}}+30x-20=0\]?

Answer 1

Hint: The given equation \[-10{{x}^{2}}+30x-20=0\] is a quadratic equation which firstly can be simplified by dividing it by the factor $-10$ which is common to all the terms on the LHS. In doing so, our equation will get reduced to ${{x}^{2}}-3x+2=0$. Then we have to use the middle term splitting method to solve it. For this, we need to break the middle term, which is $-3x$, into the sum of two terms whose product is equal to the product is equal to the product of the first term ${{x}^{2}}$ and the last term $2$, that is, the product must be equal to $2{{x}^{2}}$. So we will break the middle term as $-3x=-2x-x$ and then take factors common from the terms to factorize the LHS and finally solve it.

Complete step by step solution:
The equation given is
\[\Rightarrow -10{{x}^{2}}+30x-20=0\]
The factor of $-10$ is common to all of the three terms on the LHS which is making the equation complex. So we divide the above equation by $-10$ to get
\[\begin{align}
  & \Rightarrow \dfrac{1}{-10}\left( -10{{x}^{2}}+30x-20 \right)=\dfrac{0}{-10} \\
 & \Rightarrow {{x}^{2}}-3x+2=0 \\
\end{align}\]
Now, using the middle term splitting technique, we split the middle term as $-3x=-2x-x$ to get
\[\Rightarrow {{x}^{2}}-2x-x+2=0\]
Taking $x$ and $-1$ common from the first and the second pair of terms respectively, we get
$\Rightarrow x\left( x-2 \right)-1\left( x-2 \right)=0$
Since $\left( x-2 \right)$ is common, we can take it outside to get
$\Rightarrow \left( x-2 \right)\left( x-1 \right)=0$
Since the product of $\left( x-2 \right)$ and $\left( x-1 \right)$ is equal to zero, we can equate each of them to zero to get
\[\begin{align}
  & \Rightarrow x-2=0 \\
 & \Rightarrow x=2 \\
\end{align}\]
And
$\begin{align}
  & \Rightarrow x-1=0 \\
 & \Rightarrow x=1 \\
\end{align}$

Hence, the solutions of the given equation are $x=1$ and $x=2$.

Note: This question can also be easily solved by using the quadratic formula given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. But we have solved this question using the middle term splitting method since the product of the first and the last terms, equal to $2{{x}^{2}}$, was small and therefore the splitting of the middle term was easy. If in a quadratic equation, the product is greater, then the splitting becomes difficult and we can use the quadratic formula.