Question

The minimum value of the expression $7-20x+11{{x}^{2}}$ is:
A. $\dfrac{177}{11}$
B. $-\dfrac{177}{11}$
C. $-\dfrac{23}{11}$
D. $\dfrac{23}{11}$

Answer 1

Hint: We will do this question by using completing the square method. We will form the whole square of the given expression and it will come out as a whole square plus or minus another number. Then we will keep that value of x in this expression such that the value inside the whole square is 0. The term which is added or subtracted with the whole square term along with the sign hence will be our answer.

Complete step by step answer:
Here, we have to find the minimum value of $7-20x+11{{x}^{2}}$. We will do this by using completing the square method.
Here, the given equation is:
$7-20x+11{{x}^{2}}$
Now, rearranging the terms, we can also write it as:
$11{{x}^{2}}-20x+7$
Now, taking 11 common from this expression we get:
$\begin{align}
  & 11{{x}^{2}}-20x+7 \\
 & \Rightarrow 11\left( {{x}^{2}}-\dfrac{20}{11}x+\dfrac{7}{11} \right) \\
\end{align}$
Now, we can also write the coefficient of ‘x’ as:
$\begin{align}
  & 11\left( {{x}^{2}}-\dfrac{20}{11}x+\dfrac{7}{11} \right) \\
 & \Rightarrow 11\left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+\dfrac{7}{11} \right) \\
\end{align}$
Now, adding and subtracting the square of $\dfrac{10}{11}$ in this expression we get:
$\begin{align}
  & 11\left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+\dfrac{7}{11} \right) \\
 & \Rightarrow 11\left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+\dfrac{7}{11}+\dfrac{100}{121}-\dfrac{100}{121} \right) \\
\end{align}$
Now, rearranging these terms, we get:
$\begin{align}
  & 11\left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+\dfrac{7}{11}+\dfrac{100}{121}-\dfrac{100}{121} \right) \\
 & \Rightarrow 11\left( \left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+\dfrac{100}{121} \right)+\dfrac{7}{11}-\dfrac{100}{121} \right) \\
 & \Rightarrow 11\left( \left( {{x}^{2}}-2\left( \dfrac{10}{11} \right)\left( x \right)+{{\left( \dfrac{10}{11} \right)}^{2}} \right)-\dfrac{23}{121} \right) \\
 & \Rightarrow 11\left( {{\left( x-\dfrac{10}{11} \right)}^{2}}-\dfrac{23}{121} \right) \\
\end{align}$
Thus, the final expression comes out as:
$\begin{align}
  & 11\left( {{\left( x-\dfrac{10}{11} \right)}^{2}}-\dfrac{23}{121} \right) \\
 & \Rightarrow 11{{\left( x-\dfrac{10}{11} \right)}^{2}}-\dfrac{23}{11} \\
\end{align}$
Now, if we know that the minimum value of this expression is when the value of the whole square is 0. Hence, the minimum value of this expression is at $x=\dfrac{10}{11}$. Hence, the minimum value of the given expression is:
$\begin{align}
  & 11{{\left( x-\dfrac{10}{11} \right)}^{2}}-\dfrac{23}{11} \\
 & \Rightarrow 11{{\left( \dfrac{10}{11}-\dfrac{10}{11} \right)}^{2}}-\dfrac{23}{11} \\
 & \Rightarrow 11\left( 0 \right)-\dfrac{23}{11} \\
 & \therefore -\dfrac{23}{11} \\
\end{align}$
Hence, the minimum value of the given expression is $-\dfrac{23}{11}$.

So, the correct answer is “Option C”.

Note: We should always make the coefficient of ${{x}^{2}}$ as 1. This is because the given coefficient may or may not be a perfect square and if it isn’t a perfect square as in this case, the terms will come in the form of under root which will lead to a complicated calculation. Hence, it is always preferable to make the coefficient of ${{x}^{2}}$ as 1.