Solve 20x^2-x-1=0

$$20 { x }^{ 2 } -x-1 = 0$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=20\left(-1\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite 20x^{2}-x-1 as \left(20x^{2}-5x\right)+\left(4x-1\right).

$$\left(20x^{2}-5x\right)+\left(4x-1\right)$$

Factor out 5x in 20x^{2}-5x.

$$5x\left(4x-1\right)+4x-1$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(5x+1\right)$$

To find equation solutions, solve 4x-1=0 and 5x+1=0.

$$x=\frac{1}{4} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

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.

$$20x^{2}-x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 20\left(-1\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-1\right)±\sqrt{1-80\left(-1\right)}}{2\times 20}$$

Multiply -80 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 20}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\times 20}$$

Take the square root of 81.

$$x=\frac{-\left(-1\right)±9}{2\times 20}$$

The opposite of -1 is 1.

$$x=\frac{1±9}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{1±9}{40}$$

Now solve the equation x=\frac{1±9}{40} when ± is plus. Add 1 to 9.

$$x=\frac{10}{40}$$

Reduce the fraction \frac{10}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{1±9}{40} when ± is minus. Subtract 9 from 1.

$$x=-\frac{8}{40}$$

Reduce the fraction \frac{-8}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{1}{5}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$20x^{2}-x-1=0$$

Add 1 to both sides of the equation.

$$20x^{2}-x-1-\left(-1\right)=-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$20x^{2}-x=-\left(-1\right)$$

Subtract -1 from 0.

$$20x^{2}-x=1$$

Divide both sides by 20.

$$\frac{20x^{2}-x}{20}=\frac{1}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}-\frac{1}{20}x=\frac{1}{20}$$

Divide -\frac{1}{20}, the coefficient of the x term, by 2 to get -\frac{1}{40}. Then add the square of -\frac{1}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{1}{20}+\frac{1}{1600}$$

Add \frac{1}{20} to \frac{1}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{81}{1600}$$

Factor x^{2}-\frac{1}{20}x+\frac{1}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{40}\right)^{2}=\frac{81}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{40}\right)^{2}}=\sqrt{\frac{81}{1600}}$$

Simplify.

$$x-\frac{1}{40}=\frac{9}{40} x-\frac{1}{40}=-\frac{9}{40}$$

Add \frac{1}{40} to both sides of the equation.

$$x=\frac{1}{4} x=-\frac{1}{5}$$