15x^2-347.88x+20063.4=0

asked by guest
on Jul 06, 2024 at 6:17 am



You asked:

Solve the equation \(15 \cdot {x}^{2} - 347.88 x + 20063.4 = 0\) for the variable \(x\).

MathBot Answer:

\[\begin{aligned}x &= \frac{2899}{250} + \frac{3 i \sqrt{8354811}}{250} \approx 11.596 + 34.685628 i\\x &= \frac{2899}{250} - \frac{3 i \sqrt{8354811}}{250} \approx 11.596 - 34.685628 i\end{aligned}\]

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 \(a=15\), \(b=- \frac{8697}{25}\), and \(c=\frac{100317}{5}\).

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:\[b^{2}-4ac = \left(- \frac{8697}{25}\right)^{2} - 60 \cdot \frac{100317}{5}=- \frac{676739691}{625} < 0\] The discriminant is less than zero, so this quadratic equation has two solutions with non-zero imaginary part.

The two solutions are: \[x = \frac{\left(-1\right) \left(- \frac{8697}{25}\right) + \sqrt{- \frac{676739691}{625}}}{2 \cdot 15} = \frac{2899}{250} + \frac{3 \sqrt{8354811} i}{250} \approx 11.596 + 34.685628 i\] \[x = \frac{\left(-1\right) \left(- \frac{8697}{25}\right) - \sqrt{- \frac{676739691}{625}}}{2 \cdot 15} = \frac{2899}{250} - \frac{3 \sqrt{8354811} i}{250} \approx 11.596 -34.685628 i\]

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


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