Question

Saw the following quadrilateral equation using the perfect square trinormal pattern x^2+ x + 0.25 = 0

Answer 1

To analyze the equation \( x^2 + x + 0.25 = 0 \), we notice that it can be expressed as a perfect square trinomial.

The equation can be rewritten by recognizing that \( 0.25 \) is equal to \( \left(\frac{1}{2}\right)^2 \). Thus, we can rewrite the quadratic equation as:

\[
x^2 + x + \left(\frac{1}{2}\right)^2 = 0
\]

This fits the pattern of a perfect square trinomial, which is given by the formula \( (a + b)^2 = a^2 + 2ab + b^2 \).

In this case, we can express it as:

\[
\left(x + \frac{1}{2}\right)^2 = 0
\]

Now, we can solve this equation by taking the square root of both sides:

\[
x + \frac{1}{2} = 0
\]

Subtracting \( \frac{1}{2} \) from both sides gives:

\[
x = -\frac{1}{2}
\]

Therefore, the solution to the equation \( x^2 + x + 0.25 = 0 \) is:

\[
x = -\frac{1}{2}
\]

Since this is a perfect square equation, it has a double root, meaning \( x = -\frac{1}{2} \) is the only solution, and it occurs with multiplicity 2.