x^2-250x+25*25=0

asked by guest
on Jan 22, 2025 at 5:09 am



You asked:

Solve the equation \({x}^{2} - 250 x + 25 \cdot 25 = 0\) for the variable \(x\).

MathBot Answer:

\[\begin{aligned}x &= 125 + 50 \sqrt{6} \approx 247.47449\\x &= 125 - 50 \sqrt{6} \approx 2.5255129\end{aligned}\]

Solve by quadratic formula:

Simplify the left hand side of the equation into standard form: $$\begin{aligned}x^{2} - 250 x + 25 \cdot 25 &= 0\\x^{2} - 250 x + 625 &= 0\end{aligned}$$

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 \(a=1\), \(b=-250\), and \(c=625\).

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:\[b^{2}-4ac = \left(-250\right)^{2} - 4 \cdot 625=60000 > 0\] The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: \[x = \frac{\left(-1\right) \left(-250\right) + \sqrt{60000}}{2 \cdot 1} = 125 + 50 \sqrt{6} \approx 247.47449\] \[x = \frac{\left(-1\right) \left(-250\right) - \sqrt{60000}}{2 \cdot 1} = 125 - 50 \sqrt{6} \approx 2.5255129\]

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