$x^{2}$ +16x+336=0

asked by guest
on Nov 22, 2024 at 8:33 am



You asked:

Solve the equation \({x}^{2} + 16 x + 336 = 0\) for the variable \(x\).

MathBot Answer:

\[\begin{aligned}x &= -8 + 4 i \sqrt{17} \approx -8.0 + 16.492422 i\\x &= -8 - 4 i \sqrt{17} \approx -8.0 - 16.492422 i\end{aligned}\]

Solve by quadratic formula:

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 \(a=1\), \(b=16\), and \(c=336\).

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:\[b^{2}-4ac = 16^{2} - 4 \cdot 336=-1088 < 0\] The discriminant is less than zero, so this quadratic equation has two solutions with non-zero imaginary part.

The two solutions are: \[x = \frac{\left(-1\right) 16 + \sqrt{-1088}}{2 \cdot 1} = -8 + 4 \sqrt{17} i = -8 + 16.492422 i\] \[x = \frac{\left(-1\right) 16 - \sqrt{-1088}}{2 \cdot 1} = -8 - 4 \sqrt{17} i = -8 -16.492422 i\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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