Solve 4y^2-6y+8

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

Factor

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

Solution Steps

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

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

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.

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

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.

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

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-6\right)±\sqrt{36-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$y=\frac{-\left(-6\right)±\sqrt{36-128}}{2\times 4}$$

Add 36 to -128.

$$y=\frac{-\left(-6\right)±\sqrt{-92}}{2\times 4}$$

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

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

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

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

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

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

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

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = 2$$

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

$$\frac{9}{16} - u^2 = 2$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = 2-\frac{9}{16} = \frac{23}{16}$$

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

$$u^2 = -\frac{23}{16} u = \pm\sqrt{-\frac{23}{16}} = \pm \frac{\sqrt{23}}{4}i$$

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

$$r =\frac{3}{4} - \frac{\sqrt{23}}{4}i = 0.750 - 1.199i s = \frac{3}{4} + \frac{\sqrt{23}}{4}i = 0.750 + 1.199i$$

Evaluate

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