640\ \cdot\left[\left(1+\frac{x}{100}\right)^{2}-1\right]=\frac{77440}{100}

asked by guest
on Nov 16, 2024 at 12:35 am



You asked:

Solve the equation \(640 \cdot \left({\left( 1 + \frac{x}{100} \right)}^{2} - 1\right) = \frac{77440}{100}\) for the variable \(x\).

MathBot Answer:

\[\begin{aligned}x &= -100 + 10 \sqrt{221} \approx 48.660687\\x &= -100 - 10 \sqrt{221} \approx -248.66069\end{aligned}\]

Solve by quadratic formula:

Subtract the right hand side from the left hand side of: \[640 \cdot \left({\left( 1 + \frac{x}{100} \right)}^{2} - 1\right) = \frac{77440}{100}\] The result is a quadratic equation: \[\frac{8 x^{2}}{125} + \frac{64 x}{5} - \frac{3872}{5} = 0\]

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 \(a=\frac{8}{125}\), \(b=\frac{64}{5}\), and \(c=- \frac{3872}{5}\).

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:\[b^{2}-4ac = \left(\frac{64}{5}\right)^{2} - \frac{32}{125} \left(- \frac{3872}{5}\right)=\frac{226304}{625} > 0\] The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: \[x = \frac{\left(-1\right) \frac{64}{5} + \sqrt{\frac{226304}{625}}}{2 \cdot \frac{8}{125}} = -100 + 10 \sqrt{221} \approx 48.660687\] \[x = \frac{\left(-1\right) \frac{64}{5} - \sqrt{\frac{226304}{625}}}{2 \cdot \frac{8}{125}} = -100 - 10 \sqrt{221} \approx -248.66069\]

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