The square of a certain number is 22 less than 13 tomes the original number. Find the number

Answer 1

Let the unknown number be $x$. According to the problem, the square of the number is 22 less than 13 times the number. We can set up the equation as follows:

$$x^2=13x-22$$

Now, rearranging the equation gives us:

$$x^2-13x+22=0$$

Next, we can solve this quadratic equation using the quadratic formula:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

where $a=1$, $b=-13$, and $c=22$.

First, we calculate the discriminant:

$$b^2-4ac=(-13{)}^2-4\cdot 1\cdot 22=169-88=81$$

Since the discriminant is positive, we will have two real solutions. Now we substitute back into the quadratic formula:

$$x=\frac{-(-13)\pm \sqrt{81}}{2\cdot 1}=\frac{13\pm 9}{2}$$

This gives us two potential solutions:

1. $x=\frac{13+9}{2}=\frac{22}{2}=11$
2. $x=\frac{13-9}{2}=\frac{4}{2}=2$

Thus, the possible values of $x$ are $11$ and $2$.

To confirm, we can check both values:

1. For $x=11$: $$x^2={11}^2=121$$ $$13x-22=13\cdot 11-22=143-22=121$$ This is correct.

2. For $x=2$: $$x^2=2^2=4$$ $$13x-22=13\cdot 2-22=26-22=4$$ This is also correct.

Thus, the numbers that satisfy the condition are $\boxed{{\displaystyle 11}}$ and $\boxed{{\displaystyle 2}}$.