Solve x^2-4x+4quadx^2-4quad3x+6

$$x ^ { 2 } - 4 x + 4 \quad x ^ { 2 } - 4 \quad 3 x + 6$$

Evaluate

$$5x^{2}-16x+6$$

Solution Steps

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-4x-4\times 3x+6$$

Multiply 4 and 3 to get 12.

$$5x^{2}-4x-12x+6$$

Combine -4x and -12x to get -16x.

$$5x^{2}-16x+6$$

Factor

$$5\left(x-\frac{8-\sqrt{34}}{5}\right)\left(x-\frac{\sqrt{34}+8}{5}\right)$$

Steps Using the Quadratic Formula

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$factor(5x^{2}-4x-4\times 3x+6)$$

Multiply 4 and 3 to get 12.

$$factor(5x^{2}-4x-12x+6)$$

Combine -4x and -12x to get -16x.

$$factor(5x^{2}-16x+6)$$

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.

$$5x^{2}-16x+6=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 6}}{2\times 5}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$x=\frac{-\left(-16\right)±\sqrt{256-120}}{2\times 5}$$

Add 256 to -120.

$$x=\frac{-\left(-16\right)±\sqrt{136}}{2\times 5}$$

Take the square root of 136.

$$x=\frac{-\left(-16\right)±2\sqrt{34}}{2\times 5}$$

The opposite of -16 is 16.

$$x=\frac{16±2\sqrt{34}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{16±2\sqrt{34}}{10}$$

Now solve the equation x=\frac{16±2\sqrt{34}}{10} when ± is plus. Add 16 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}+16}{10}$$

Divide 16+2\sqrt{34} by 10.

$$x=\frac{\sqrt{34}+8}{5}$$

Now solve the equation x=\frac{16±2\sqrt{34}}{10} when ± is minus. Subtract 2\sqrt{34} from 16.

$$x=\frac{16-2\sqrt{34}}{10}$$

Divide 16-2\sqrt{34} by 10.

$$x=\frac{8-\sqrt{34}}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8+\sqrt{34}}{5} for x_{1} and \frac{8-\sqrt{34}}{5} for x_{2}.

$$5x^{2}-16x+6=5\left(x-\frac{\sqrt{34}+8}{5}\right)\left(x-\frac{8-\sqrt{34}}{5}\right)$$