Solve x^2+25x-125

$$x ^ { 2 } + 25 x - 125$$

Factor

$$\left(x-\frac{-15\sqrt{5}-25}{2}\right)\left(x-\frac{15\sqrt{5}-25}{2}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+25x-125=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-125\right)}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-125\right)}}{2}$$

Multiply -4 times -125.

$$x=\frac{-25±\sqrt{625+500}}{2}$$

Add 625 to 500.

$$x=\frac{-25±\sqrt{1125}}{2}$$

Take the square root of 1125.

$$x=\frac{-25±15\sqrt{5}}{2}$$

Now solve the equation x=\frac{-25±15\sqrt{5}}{2} when ± is plus. Add -25 to 15\sqrt{5}.

$$x=\frac{15\sqrt{5}-25}{2}$$

Now solve the equation x=\frac{-25±15\sqrt{5}}{2} when ± is minus. Subtract 15\sqrt{5} from -25.

$$x=\frac{-15\sqrt{5}-25}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-25+15\sqrt{5}}{2} for x_{1} and \frac{-25-15\sqrt{5}}{2} for x_{2}.

$$x^{2}+25x-125=\left(x-\frac{15\sqrt{5}-25}{2}\right)\left(x-\frac{-15\sqrt{5}-25}{2}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +25x -125 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -25 rs = -125$$

Two numbers r and s sum up to -25 exactly when the average of the two numbers is \frac{1}{2}*-25 = -\frac{25}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{25}{2} - u s = -\frac{25}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -125

$$(-\frac{25}{2} - u) (-\frac{25}{2} + u) = -125$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{625}{4} - u^2 = -125$$

Simplify the expression by subtracting \frac{625}{4} on both sides

$$-u^2 = -125-\frac{625}{4} = -\frac{1125}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1125}{4} u = \pm\sqrt{\frac{1125}{4}} = \pm \frac{\sqrt{1125}}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{25}{2} - \frac{\sqrt{1125}}{2} = -29.271 s = -\frac{25}{2} + \frac{\sqrt{1125}}{2} = 4.271$$

Evaluate

$$x^{2}+25x-125$$