x^2+39=23 Solve for x
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Solve by quadratic formula:
Subtract the right hand side from the left hand side of: \[{x}^{2} + 39 = 23\] The result is a quadratic equation: \[x^{2} + 16 = 0\]
Given a quadratic equation \(a x^{2} + b x + c = 0\), where \(a\), \(b\), \(c\) are constants and \(a \ne 0\), the solutions are given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] In this case \(a=1\), \(b=0\), and \(c=16\).
The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:\[b^{2}-4ac = 0^{2} - 4 \cdot 16=-64 < 0\] The discriminant is less than zero, so this quadratic equation has two solutions with non-zero imaginary part.
The two solutions are: \[x = \frac{\left(-1\right) 0 + \sqrt{-64}}{2 \cdot 1} = 4 i \] \[x = \frac{\left(-1\right) 0 - \sqrt{-64}}{2 \cdot 1} = -4 i \]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).
\(i\) is the imaginary unit, defined as \(i^2 = -1\).