Question

1. solve each equation algebraically.

a) $2x^2 + 98 = 0$

Explanation:

Step1: Isolate the \(x^2\) term

Subtract 98 from both sides of the equation \(2x^2 + 98 = 0\):
\(2x^2 = -98\)

Step2: Solve for \(x^2\)

Divide both sides by 2:
\(x^2 = \frac{-98}{2} = -49\)

Step3: Analyze the square root of a negative number

The square root of a negative number is not a real number (in the set of real numbers). For real solutions, since \(x^2 = -49\) has no real solutions (because the square of any real number \(x\) is non - negative, i.e., \(x^2\geq0\) for all real \(x\)), and if we consider complex numbers, we can write \(x=\pm\sqrt{- 49}=\pm7i\) (where \(i = \sqrt{-1}\) is the imaginary unit). But if we are restricted to real numbers, there is no real solution.

Answer:

In the set of real numbers, there is no solution. In the set of complex numbers, \(x = \pm7i\)