Question

Four times the square of a positive number decreased by $6$ equals $7$times the number increased by $9$. Find the number.

Answer 1

Step 1: Determining the quadratic equation in $x$

Given that, the four times the square of a positive number decreased by $6$ equals $7$times the number increased by $9$.

Let the positive number be $x$.

Therefore, we have a quadratic equation

$$\begin{gathered} \Rightarrow 4x^2-6=7x+9 \\ \Rightarrow 4x^2-7x-15=0 \end{gathered}$$

Step 2: Determining the positive number $\left(x\right)$

By using the factorization method

$$\begin{gathered} \Rightarrow 4x^2-7x-15=0 \\ \Rightarrow 4x-12x+5x-15=0 \\ \Rightarrow \left(x-3\right)\left(4x-5\right)=0 \\ \Rightarrow x=-\frac{5}{4},3 \end{gathered}$$

Here the negative value of $x$ is not considerable. So, $x=3$

Hence, the number is $3$.