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

1 answer

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 \( x = 2 \):

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

  1. 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 \( \boxed{-1} \).