Question

How do you solve ${{x}^{2}}-10x+21\le 0$?

Answer 1

Hint: To solve the given inequality, we are going to first of all factorize the quadratic equation given on the L.H.S of the above inequality which we are going to do by multiplying the coefficient of ${{x}^{2}}$ and the constant and then write the factors for this result of multiplication. And then arrange the factors in such a way so that addition or subtraction will give the coefficient of x. After that we will find the solutions for the inequality.

Complete step-by-step solution:
The inequality given above which we have to solve is as follows:
${{x}^{2}}-10x+21\le 0$
As you can see that the highest power of this equation is 2 so the degree of the above equation is 2 and hence, the L.H.S of the above inequality is quadratic.
Now, to solve the above quadratic expression ${{x}^{2}}-10x+21$ , we are going to factorize this expression by finding the factors of 21.
$\begin{align}
  & 21=1\times 21 \\
 & 21=3\times 7 \\
\end{align}$
If you look at the last factor then you will find that on adding this factor (3 + 7) you will get the coefficient of x which is 10 so substituting $\left( 3+7 \right)$ in place of 10 in the above equation we get:
$\begin{align}
  & {{x}^{2}}-\left( 3+7 \right)x+21 \\
 & \Rightarrow {{x}^{2}}-3x-7x+21 \\
\end{align}$
Taking x as common from first two terms in the above expression and -7 as common from the last two terms and we get,
$x\left( x-3 \right)-7\left( x-3 \right)$
Now, taking $\left( x-3 \right)$ as common from the above we get,
$\left( x-3 \right)\left( x-7 \right)$
 Now, putting the inequality in the above expression we get,
$\left( x-3 \right)\left( x-7 \right)\le 0$
The critical points of the above expression are 3 and 7 so when x is less than 3 then we get an overall positive solution because both the expressions $\left( x-3 \right)\And \left( x-7 \right)$ will be negative and their multiplication becomes positive. And when x is greater than 7 so both $\left( x-3 \right)\And \left( x-7 \right)$ will be positive and overall multiplication is positive. But we require the multiplication to be negative and equal to 0 which would be possible when x lies either between 3 and 7 or when x takes value as 3 and 7 so the solution for the above inequality is as follows:
$x\in \left[ 3,7 \right]$
Hence, we have found the solution of the given inequality as follows: $x\in \left[ 3,7 \right]$


Note: The mistake that could be possible in the above solution is that either applying closed bracket in the final solution you may use the open brackets which will make the solution look like as follows:
$x\in \left( 3,7 \right)$
This is the wrong solution because the open bracket is showing that we have not taken into account the solutions 3 and 7 so beware to make such mistakes in the examination.