The square of a number is added to 37 equal to the difference of a number and one-eighths.
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MathBot Answer:
Solve by quadratic formula:
Subtract the right hand side from the left hand side of: \[{\left( a number + 37 \right)}^{2} = a number - \frac{1}{8}\] The result is a quadratic equation: \[a number^{2} + 73 a number + \frac{10953}{8} = 0\]
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 \(a=1\), \(b=73\), and \(c=\frac{10953}{8}\).
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:\[b^{2}-4ac = 73^{2} - 4 \cdot \frac{10953}{8}=- \frac{295}{2} < 0\] The discriminant is less than zero, so this quadratic equation has two solutions with non-zero imaginary part.
The two solutions are: \[a number = \frac{\left(-1\right) 73 + \sqrt{- \frac{295}{2}}}{2 \cdot 1} = - \frac{73}{2} + \frac{\sqrt{590} i}{4} \approx -36.5 + 6.0724789 i\] \[a number = \frac{\left(-1\right) 73 - \sqrt{- \frac{295}{2}}}{2 \cdot 1} = - \frac{73}{2} - \frac{\sqrt{590} i}{4} \approx -36.5 -6.0724789 i\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).
\(i\) is the imaginary unit, defined as \(i^2 = -1\).