Solve 4t^2-8t+6

$$4 t ^ { 2 } - 8 t + 6$$

Factor

$$2\left(2t^{2}-4t+3\right)$$

Solution Steps

Factor out 2. Polynomial 2t^{2}-4t+3 is not factored since it does not have any rational roots.

$$2\left(2t^{2}-4t+3\right)$$

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.

$$4t^{2}-8t+6=0$$

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.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Square -8.

$$t=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-\left(-8\right)±\sqrt{64-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$t=\frac{-\left(-8\right)±\sqrt{64-96}}{2\times 4}$$

Add 64 to -96.

$$t=\frac{-\left(-8\right)±\sqrt{-32}}{2\times 4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$4t^{2}-8t+6$$

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

$$x ^ 2 -2x +\frac{3}{2} = 0$$

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

$$r + s = 2 rs = \frac{3}{2}$$

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{3}{2}

$$(1 - u) (1 + u) = \frac{3}{2}$$

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

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

Simplify the expression by subtracting 1 on both sides

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

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

$$u^2 = -\frac{1}{2} u = \pm\sqrt{-\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}i$$

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

$$r =1 - \frac{1}{\sqrt{2}}i = 1 - 0.707i s = 1 + \frac{1}{\sqrt{2}}i = 1 + 0.707i$$

Evaluate

$$4t^{2}-8t+6$$