Question

how many real solutions does the equation have?
$x^2 + 63 = -29$
no real solution
one real solution
two real solutions

Explanation:

Step1: Analyze the equation \(x^2 + 63 = -29\)

We can rewrite the equation to solve for \(x^2\) by subtracting 63 from both sides. So, \(x^2=-29 - 63\).

Step2: Calculate the right - hand side

\(-29-63=-92\), so the equation becomes \(x^2=-92\).
In the set of real numbers, the square of any real number \(x\) (i.e., \(x^2\)) is always non - negative (because if \(x\) is positive, \(x^2=x\times x>0\); if \(x = 0\), \(x^2 = 0\); if \(x\) is negative, \(x^2=(-x)\times(-x)>0\)). But here we have \(x^2=-92\), and \(-92<0\). So, there are no real numbers \(x\) that satisfy this equation.

Answer:

no real solution