Solve 5y^2-26y+9

$$5 y ^ { 2 } - 26 y + 9$$

Factor

$$5\left(y-\frac{13-2\sqrt{31}}{5}\right)\left(y-\frac{2\sqrt{31}+13}{5}\right)$$

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.

$$5y^{2}-26y+9=0$$

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.

$$y=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 5\times 9}}{2\times 5}$$

Square -26.

$$y=\frac{-\left(-26\right)±\sqrt{676-4\times 5\times 9}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-26\right)±\sqrt{676-20\times 9}}{2\times 5}$$

Multiply -20 times 9.

$$y=\frac{-\left(-26\right)±\sqrt{676-180}}{2\times 5}$$

Add 676 to -180.

$$y=\frac{-\left(-26\right)±\sqrt{496}}{2\times 5}$$

Take the square root of 496.

$$y=\frac{-\left(-26\right)±4\sqrt{31}}{2\times 5}$$

The opposite of -26 is 26.

$$y=\frac{26±4\sqrt{31}}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{26±4\sqrt{31}}{10}$$

Now solve the equation y=\frac{26±4\sqrt{31}}{10} when ± is plus. Add 26 to 4\sqrt{31}.

$$y=\frac{4\sqrt{31}+26}{10}$$

Divide 26+4\sqrt{31} by 10.

$$y=\frac{2\sqrt{31}+13}{5}$$

Now solve the equation y=\frac{26±4\sqrt{31}}{10} when ± is minus. Subtract 4\sqrt{31} from 26.

$$y=\frac{26-4\sqrt{31}}{10}$$

Divide 26-4\sqrt{31} by 10.

$$y=\frac{13-2\sqrt{31}}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13+2\sqrt{31}}{5} for x_{1} and \frac{13-2\sqrt{31}}{5} for x_{2}.

$$5y^{2}-26y+9=5\left(y-\frac{2\sqrt{31}+13}{5}\right)\left(y-\frac{13-2\sqrt{31}}{5}\right)$$

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

$$x ^ 2 -\frac{26}{5}x +\frac{9}{5} = 0$$

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

$$r + s = \frac{26}{5} rs = \frac{9}{5}$$

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-gzdabgg4ehffg0hf.b01.azurefd.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{9}{5}

$$(\frac{13}{5} - u) (\frac{13}{5} + u) = \frac{9}{5}$$

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

$$\frac{169}{25} - u^2 = \frac{9}{5}$$

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

$$-u^2 = \frac{9}{5}-\frac{169}{25} = -\frac{124}{25}$$

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

$$u^2 = \frac{124}{25} u = \pm\sqrt{\frac{124}{25}} = \pm \frac{\sqrt{124}}{5}$$

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

$$r =\frac{13}{5} - \frac{\sqrt{124}}{5} = 0.373 s = \frac{13}{5} + \frac{\sqrt{124}}{5} = 4.827$$

Evaluate

$$5y^{2}-26y+9$$