Solve y^2+8y+4)

$$y ^ { 2 } + 8 y + 4 )$$

Factor

$$\left(y-\left(-2\sqrt{3}-4\right)\right)\left(y-\left(2\sqrt{3}-4\right)\right)$$

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.

$$y^{2}+8y+4=0$$

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.

$$y=\frac{-8±\sqrt{8^{2}-4\times 4}}{2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 4}}{2}$$

Multiply -4 times 4.

$$y=\frac{-8±\sqrt{64-16}}{2}$$

Add 64 to -16.

$$y=\frac{-8±\sqrt{48}}{2}$$

Take the square root of 48.

$$y=\frac{-8±4\sqrt{3}}{2}$$

Now solve the equation y=\frac{-8±4\sqrt{3}}{2} when ± is plus. Add -8 to 4\sqrt{3}.

$$y=\frac{4\sqrt{3}-8}{2}$$

Divide -8+4\sqrt{3} by 2.

$$y=2\sqrt{3}-4$$

Now solve the equation y=\frac{-8±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from -8.

$$y=\frac{-4\sqrt{3}-8}{2}$$

Divide -8-4\sqrt{3} by 2.

$$y=-2\sqrt{3}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+2\sqrt{3} for x_{1} and -4-2\sqrt{3} for x_{2}.

$$y^{2}+8y+4=\left(y-\left(2\sqrt{3}-4\right)\right)\left(y-\left(-2\sqrt{3}-4\right)\right)$$

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.

$$x ^ 2 +8x +4 = 0$$

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

$$r + s = -8 rs = 4$$

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 = 4

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

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

$$16 - u^2 = 4$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 4-16 = -12$$

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

$$u^2 = 12 u = \pm\sqrt{12} = \pm \sqrt{12}$$

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

$$r =-4 - \sqrt{12} = -7.464 s = -4 + \sqrt{12} = -0.536$$

Evaluate

$$y^{2}+8y+4$$