Solve 5u^2+40u+50

$$5 u ^ { 2 } + 40 u + 50$$

Factor

$$5\left(u-\left(-\sqrt{6}-4\right)\right)\left(u-\left(\sqrt{6}-4\right)\right)$$

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.

$$5u^{2}+40u+50=0$$

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.

$$u=\frac{-40±\sqrt{40^{2}-4\times 5\times 50}}{2\times 5}$$

Square 40.

$$u=\frac{-40±\sqrt{1600-4\times 5\times 50}}{2\times 5}$$

Multiply -4 times 5.

$$u=\frac{-40±\sqrt{1600-20\times 50}}{2\times 5}$$

Multiply -20 times 50.

$$u=\frac{-40±\sqrt{1600-1000}}{2\times 5}$$

Add 1600 to -1000.

$$u=\frac{-40±\sqrt{600}}{2\times 5}$$

Take the square root of 600.

$$u=\frac{-40±10\sqrt{6}}{2\times 5}$$

Multiply 2 times 5.

$$u=\frac{-40±10\sqrt{6}}{10}$$

Now solve the equation u=\frac{-40±10\sqrt{6}}{10} when ± is plus. Add -40 to 10\sqrt{6}.

$$u=\frac{10\sqrt{6}-40}{10}$$

Divide -40+10\sqrt{6} by 10.

$$u=\sqrt{6}-4$$

Now solve the equation u=\frac{-40±10\sqrt{6}}{10} when ± is minus. Subtract 10\sqrt{6} from -40.

$$u=\frac{-10\sqrt{6}-40}{10}$$

Divide -40-10\sqrt{6} by 10.

$$u=-\sqrt{6}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+\sqrt{6} for x_{1} and -4-\sqrt{6} for x_{2}.

$$5u^{2}+40u+50=5\left(u-\left(\sqrt{6}-4\right)\right)\left(u-\left(-\sqrt{6}-4\right)\right)$$

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.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 +8x +10 = 0$$

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

$$r + s = -8 rs = 10$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(-4 - u) (-4 + u) = 10$$

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

$$16 - u^2 = 10$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 10-16 = -6$$

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

$$u^2 = 6 u = \pm\sqrt{6} = \pm \sqrt{6}$$

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

$$r =-4 - \sqrt{6} = -6.449 s = -4 + \sqrt{6} = -1.551$$

Evaluate

$$5\left(u^{2}+8u+10\right)$$