Solve 1x^2+20x+25

$$1 x ^ { 2 } + 20 x + 25$$

Factor

$$\left(x-\left(-5\sqrt{3}-10\right)\right)\left(x-\left(5\sqrt{3}-10\right)\right)$$

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.

$$x^{2}+20x+25=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 25}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 25}}{2}$$

Multiply -4 times 25.

$$x=\frac{-20±\sqrt{400-100}}{2}$$

Add 400 to -100.

$$x=\frac{-20±\sqrt{300}}{2}$$

Take the square root of 300.

$$x=\frac{-20±10\sqrt{3}}{2}$$

Now solve the equation x=\frac{-20±10\sqrt{3}}{2} when ± is plus. Add -20 to 10\sqrt{3}.

$$x=\frac{10\sqrt{3}-20}{2}$$

Divide -20+10\sqrt{3} by 2.

$$x=5\sqrt{3}-10$$

Now solve the equation x=\frac{-20±10\sqrt{3}}{2} when ± is minus. Subtract 10\sqrt{3} from -20.

$$x=\frac{-10\sqrt{3}-20}{2}$$

Divide -20-10\sqrt{3} by 2.

$$x=-5\sqrt{3}-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -10+5\sqrt{3} for x_{1} and -10-5\sqrt{3} for x_{2}.

$$x^{2}+20x+25=\left(x-\left(5\sqrt{3}-10\right)\right)\left(x-\left(-5\sqrt{3}-10\right)\right)$$

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.

$$x ^ 2 +20x +25 = 0$$

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

$$r + s = -20 rs = 25$$

Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -10. 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>

$$r = -10 - u s = -10 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 25

$$(-10 - u) (-10 + u) = 25$$

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

$$100 - u^2 = 25$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = 25-100 = -75$$

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

$$u^2 = 75 u = \pm\sqrt{75} = \pm \sqrt{75}$$

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

$$r =-10 - \sqrt{75} = -18.660 s = -10 + \sqrt{75} = -1.340$$

Evaluate

$$x^{2}+20x+25$$