Question

When $8$ is subtracted from the square of a number, the result is $7$ times the number. Find a positive solution.

Answer 1

Find a positive solution.

Step 1: Assuming the required data:

Consider that the number is $n$.

The square of the number will be $n^2$.

When $8$ is subtracted from the square of the number, then the number becomes $n^2-8$.

$7$ times the number will be $7n$.

The equation can be written as,

$\begin{array}{rcl}{n}^2-8& =& 7n\end{array}$

Step 2: Find the factors of the quadratic equation.

$\begin{array}{rcl}{n}^2-8& =& 7n\\ & \Rightarrow & n^2-8-7n=0\\ & \Rightarrow & n^2-7n-8=0\\ & \Rightarrow & n^2-8n+n-8=0\\ & \Rightarrow & n\left(n-8\right)+1\left(n-8\right)=0\\ & \Rightarrow & \left(n+1\right)\left(n-8\right)=0\end{array}$

Step 3: Equate factors to zero.

Equate each factor to zero and find the value of $n$.

$$\begin{gathered} \Rightarrow n+1=0 \\ \therefore n=-1 \\ \Rightarrow n-8=0 \\ \therefore n=8 \end{gathered}$$

As, $-1$ is the negative solution. The only positive solution is $8$.

Hence, the positive solution is $\mathbf{8}$.