Solve 10x^2-9x+1

$$10 { x }^{ 2 } -9x+1$$

Factor

$$10\left(x-\frac{9-\sqrt{41}}{20}\right)\left(x-\frac{\sqrt{41}+9}{20}\right)$$

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.

$$10x^{2}-9x+1=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10}}{2\times 10}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 10}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 10}$$

Add 81 to -40.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 10}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{41}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{9±\sqrt{41}}{20}$$

Now solve the equation x=\frac{9±\sqrt{41}}{20} when ± is plus. Add 9 to \sqrt{41}.

$$x=\frac{\sqrt{41}+9}{20}$$

Now solve the equation x=\frac{9±\sqrt{41}}{20} when ± is minus. Subtract \sqrt{41} from 9.

$$x=\frac{9-\sqrt{41}}{20}$$

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}}{20} for x_{1} and \frac{9-\sqrt{41}}{20} for x_{2}.

$$10x^{2}-9x+1=10\left(x-\frac{\sqrt{41}+9}{20}\right)\left(x-\frac{9-\sqrt{41}}{20}\right)$$

Evaluate

$$10x^{2}-9x+1$$