Solve 5z^2-26z+8

$$5 z ^ { 2 } - 26 z + 8$$

Factor

$$5\left(z-\frac{13-\sqrt{129}}{5}\right)\left(z-\frac{\sqrt{129}+13}{5}\right)$$

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.

$$5z^{2}-26z+8=0$$

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.

$$z=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 5\times 8}}{2\times 5}$$

Square -26.

$$z=\frac{-\left(-26\right)±\sqrt{676-4\times 5\times 8}}{2\times 5}$$

Multiply -4 times 5.

$$z=\frac{-\left(-26\right)±\sqrt{676-20\times 8}}{2\times 5}$$

Multiply -20 times 8.

$$z=\frac{-\left(-26\right)±\sqrt{676-160}}{2\times 5}$$

Add 676 to -160.

$$z=\frac{-\left(-26\right)±\sqrt{516}}{2\times 5}$$

Take the square root of 516.

$$z=\frac{-\left(-26\right)±2\sqrt{129}}{2\times 5}$$

The opposite of -26 is 26.

$$z=\frac{26±2\sqrt{129}}{2\times 5}$$

Multiply 2 times 5.

$$z=\frac{26±2\sqrt{129}}{10}$$

Now solve the equation z=\frac{26±2\sqrt{129}}{10} when ± is plus. Add 26 to 2\sqrt{129}.

$$z=\frac{2\sqrt{129}+26}{10}$$

Divide 26+2\sqrt{129} by 10.

$$z=\frac{\sqrt{129}+13}{5}$$

Now solve the equation z=\frac{26±2\sqrt{129}}{10} when ± is minus. Subtract 2\sqrt{129} from 26.

$$z=\frac{26-2\sqrt{129}}{10}$$

Divide 26-2\sqrt{129} by 10.

$$z=\frac{13-\sqrt{129}}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13+\sqrt{129}}{5} for x_{1} and \frac{13-\sqrt{129}}{5} for x_{2}.

$$5z^{2}-26z+8=5\left(z-\frac{\sqrt{129}+13}{5}\right)\left(z-\frac{13-\sqrt{129}}{5}\right)$$

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 5

$$x ^ 2 -\frac{26}{5}x +\frac{8}{5} = 0$$

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

$$r + s = \frac{26}{5} rs = \frac{8}{5}$$

Two numbers r and s sum up to \frac{26}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{26}{5} = \frac{13}{5}. 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>

$$r = \frac{13}{5} - u s = \frac{13}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{5}

$$(\frac{13}{5} - u) (\frac{13}{5} + u) = \frac{8}{5}$$

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

$$\frac{169}{25} - u^2 = \frac{8}{5}$$

Simplify the expression by subtracting \frac{169}{25} on both sides

$$-u^2 = \frac{8}{5}-\frac{169}{25} = -\frac{129}{25}$$

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

$$u^2 = \frac{129}{25} u = \pm\sqrt{\frac{129}{25}} = \pm \frac{\sqrt{129}}{5}$$

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

$$r =\frac{13}{5} - \frac{\sqrt{129}}{5} = 0.328 s = \frac{13}{5} + \frac{\sqrt{129}}{5} = 4.872$$

Evaluate

$$5z^{2}-26z+8$$