Question

How many real solutions does x2−4x=−4x^2-4x=-4x
2
−4x=−4 have?

Answer 1

To solve the equation \( x^2 - 4x = -4 \), we first rearrange it into a standard quadratic equation form.

Starting from the equation:

\[
x^2 - 4x + 4 = 0
\]

Next, let's simplify this:

\[
x^2 - 4x + 4 = (x - 2)^2 = 0
\]

Now, we can find the solutions by setting the squared term to zero:

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

This implies:

\[
x - 2 = 0
\]

Thus, we find that:

\[
x = 2
\]

Since the equation \( (x - 2)^2 = 0 \) has a double root, we conclude that there is exactly **one unique real solution** to the original equation.

Therefore, the number of real solutions to the equation \( x^2 - 4x = -4 \) is:

\[
\boxed{1}
\]