To determine how many real solutions exist for the equation
[tex]\[ x^2 = \frac{1}{x+3}, \][/tex]
we first manipulate the equation into a standard polynomial form. Here are the step-by-step details for solving this equation:
1. Start with the given equation:
[tex]\[ x^2 = \frac{1}{x+3}. \][/tex]
2. Eliminate the fraction by multiplying both sides by [tex]\(x+3\)[/tex]:
[tex]\[ x^2 (x + 3) = 1. \][/tex]
3. Distribute [tex]\(x^2\)[/tex] on the left-hand side:
[tex]\[ x^3 + 3x^2 = 1. \][/tex]
4. Rearrange the terms to set the equation to zero:
[tex]\[ x^3 + 3x^2 - 1 = 0. \][/tex]
Now we are left with the polynomial equation
[tex]\[ x^3 + 3x^2 - 1 = 0. \][/tex]
5. Next, we need to find the real solutions to this polynomial equation.
To do this, you can use various algebraic or numerical methods such as factoring (if possible), graphing, or using a computational tool.
Upon analyzing this polynomial equation for real solutions, one finds that:
[tex]\[ x^3 + 3x^2 - 1 = 0 \][/tex]
has no real solutions.
Therefore, there are no real numbers [tex]\( x \)[/tex] that satisfy the equation [tex]\( x^3 + 3x^2 - 1 = 0 \)[/tex].
In conclusion, the equation
[tex]\[ x^2 = \frac{1}{x+3} \][/tex]
has [tex]\(\boxed{0}\)[/tex] real solutions.
