Question

A number is two less than its square. Find all such numbers

Answer 1

Let \( x \) be the number we want to find. According to the problem, the number is two less than its square, which can be expressed with the equation:

\[
x^2 - x - 2 = 0
\]

To solve this quadratic equation, we can factor it. First, we look for two numbers that multiply to \( -2 \) (the constant term) and add to \( -1 \) (the coefficient of \( x \)). The numbers that satisfy these conditions are \( -2 \) and \( 1 \).

Thus, we can factor the equation as follows:

\[
(x - 2)(x + 1) = 0
\]

Setting each factor equal to zero gives us the possible solutions:

\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

Therefore, the numbers that are two less than their square are:

\[
\boxed{2} \quad \text{and} \quad \boxed{-1}
\]

To verify, we can substitute these values back into the original condition:

1. For \( x = 2 \):

\[
2^2 = 4 \quad \text{and} \quad 4 - 2 = 2
\]

2. For \( x = -1 \):

\[
(-1)^2 = 1 \quad \text{and} \quad 1 - 2 = -1
\]

Both conditions hold true, confirming the solutions are correct. Hence, the final answers are \( \boxed{2} \) and \( \boxed{-1} \).