Solve (8)quad5x^2+4x-1=

$$( 8 ) \quad 5 x ^ { 2 } + 4 x - 1 =$$

Factor

$$40\left(x-\frac{-\sqrt{11}-1}{20}\right)\left(x-\frac{\sqrt{11}-1}{20}\right)$$

Steps Using the Quadratic Formula

Multiply 8 and 5 to get 40.

$$factor(40x^{2}+4x-1)$$

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.

$$40x^{2}+4x-1=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 40\left(-1\right)}}{2\times 40}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 40\left(-1\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-4±\sqrt{16-160\left(-1\right)}}{2\times 40}$$

Multiply -160 times -1.

$$x=\frac{-4±\sqrt{16+160}}{2\times 40}$$

Add 16 to 160.

$$x=\frac{-4±\sqrt{176}}{2\times 40}$$

Take the square root of 176.

$$x=\frac{-4±4\sqrt{11}}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{-4±4\sqrt{11}}{80}$$

Now solve the equation x=\frac{-4±4\sqrt{11}}{80} when ± is plus. Add -4 to 4\sqrt{11}.

$$x=\frac{4\sqrt{11}-4}{80}$$

Divide -4+4\sqrt{11} by 80.

$$x=\frac{\sqrt{11}-1}{20}$$

Now solve the equation x=\frac{-4±4\sqrt{11}}{80} when ± is minus. Subtract 4\sqrt{11} from -4.

$$x=\frac{-4\sqrt{11}-4}{80}$$

Divide -4-4\sqrt{11} by 80.

$$x=\frac{-\sqrt{11}-1}{20}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{11}}{20} for x_{1} and \frac{-1-\sqrt{11}}{20} for x_{2}.

$$40x^{2}+4x-1=40\left(x-\frac{\sqrt{11}-1}{20}\right)\left(x-\frac{-\sqrt{11}-1}{20}\right)$$

Evaluate

$$40x^{2}+4x-1$$

Solution Steps

Multiply 8 and 5 to get 40.

$$40x^{2}+4x-1$$