Solve 2x^2-9x+5

$$2 x ^ { 2 } - 9 x + 5$$

Factor

$$2\left(x-\frac{9-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+9}{4}\right)$$

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.

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

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.

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

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 2}$$

Add 81 to -40.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±\sqrt{41}}{4}$$

Now solve the equation x=\frac{9±\sqrt{41}}{4} when ± is plus. Add 9 to \sqrt{41}.

$$x=\frac{\sqrt{41}+9}{4}$$

Now solve the equation x=\frac{9±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from 9.

$$x=\frac{9-\sqrt{41}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9+\sqrt{41}}{4} for x_{1} and \frac{9-\sqrt{41}}{4} for x_{2}.

$$2x^{2}-9x+5=2\left(x-\frac{\sqrt{41}+9}{4}\right)\left(x-\frac{9-\sqrt{41}}{4}\right)$$

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 2

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

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

$$r + s = \frac{9}{2} rs = \frac{5}{2}$$

Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{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>

$$r = \frac{9}{4} - u s = \frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}

$$(\frac{9}{4} - u) (\frac{9}{4} + u) = \frac{5}{2}$$

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

$$\frac{81}{16} - u^2 = \frac{5}{2}$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = \frac{5}{2}-\frac{81}{16} = -\frac{41}{16}$$

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

$$u^2 = \frac{41}{16} u = \pm\sqrt{\frac{41}{16}} = \pm \frac{\sqrt{41}}{4}$$

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

$$r =\frac{9}{4} - \frac{\sqrt{41}}{4} = 0.649 s = \frac{9}{4} + \frac{\sqrt{41}}{4} = 3.851$$

Evaluate

$$2x^{2}-9x+5$$