Solve 4t^2+8t+1

$$4 t ^ { 2 } + 8 t + 1$$

Factor

$$4\left(t-\left(-\frac{\sqrt{3}}{2}-1\right)\right)\left(t-\left(\frac{\sqrt{3}}{2}-1\right)\right)$$

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.

$$4t^{2}+8t+1=0$$

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.

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

Square 8.

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

Multiply -4 times 4.

$$t=\frac{-8±\sqrt{64-16}}{2\times 4}$$

Add 64 to -16.

$$t=\frac{-8±\sqrt{48}}{2\times 4}$$

Take the square root of 48.

$$t=\frac{-8±4\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{-8±4\sqrt{3}}{8}$$

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

$$t=\frac{4\sqrt{3}-8}{8}$$

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

$$t=\frac{\sqrt{3}}{2}-1$$

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

$$t=\frac{-4\sqrt{3}-8}{8}$$

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

$$t=-\frac{\sqrt{3}}{2}-1$$

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

$$4t^{2}+8t+1=4\left(t-\left(\frac{\sqrt{3}}{2}-1\right)\right)\left(t-\left(-\frac{\sqrt{3}}{2}-1\right)\right)$$

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 4

$$x ^ 2 +2x +\frac{1}{4} = 0$$

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

$$r + s = -2 rs = \frac{1}{4}$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}

$$(-1 - u) (-1 + u) = \frac{1}{4}$$

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

$$1 - u^2 = \frac{1}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{1}{4}-1 = -\frac{3}{4}$$

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

$$u^2 = \frac{3}{4} u = \pm\sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$$

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

$$r =-1 - \frac{\sqrt{3}}{2} = -1.866 s = -1 + \frac{\sqrt{3}}{2} = -0.134$$

Evaluate

$$4t^{2}+8t+1$$