Solve D^2+6D+4

$$D ^ { 2 } + 6 D + 4$$

Factor

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

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.

$$D^{2}+6D+4=0$$

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.

$$D=\frac{-6±\sqrt{6^{2}-4\times 4}}{2}$$

Square 6.

$$D=\frac{-6±\sqrt{36-4\times 4}}{2}$$

Multiply -4 times 4.

$$D=\frac{-6±\sqrt{36-16}}{2}$$

Add 36 to -16.

$$D=\frac{-6±\sqrt{20}}{2}$$

Take the square root of 20.

$$D=\frac{-6±2\sqrt{5}}{2}$$

Now solve the equation D=\frac{-6±2\sqrt{5}}{2} when ± is plus. Add -6 to 2\sqrt{5}.

$$D=\frac{2\sqrt{5}-6}{2}$$

Divide -6+2\sqrt{5} by 2.

$$D=\sqrt{5}-3$$

Now solve the equation D=\frac{-6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -6.

$$D=\frac{-2\sqrt{5}-6}{2}$$

Divide -6-2\sqrt{5} by 2.

$$D=-\sqrt{5}-3$$

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

$$D^{2}+6D+4=\left(D-\left(\sqrt{5}-3\right)\right)\left(D-\left(-\sqrt{5}-3\right)\right)$$

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.

$$x ^ 2 +6x +4 = 0$$

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

$$r + s = -6 rs = 4$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 4

$$(-3 - u) (-3 + u) = 4$$

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

$$9 - u^2 = 4$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 4-9 = -5$$

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

$$u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{5}$$

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

$$r =-3 - \sqrt{5} = -5.236 s = -3 + \sqrt{5} = -0.764$$

Evaluate

$$D^{2}+6D+4$$