Answer:

So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, x^1=xx

1

=xx, start superscript, 1, end superscript, equals, x.

You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides.

In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn

how to solve factored equations like (x-1)(x+3)=0(x−1)(x+3)=0left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0 and

how to use factorization methods in order to bring other equations ((left parenthesislike x^2-3x-10=0)x

2

−3x−10=0)x, squared, minus, 3, x, minus, 10, equals, 0, right parenthesis to a factored form and solve them.

Answer:

The roots of the quadratic equation \frac{1}{x}-\frac{1}{x-2} = 3

x

1

−

x−2

1

=3 are x =\frac{3+1\sqrt{3}}{3}x=

3

3+1

3

,x =\frac{3-1\sqrt{3}}{3}x=

3

3−1

3

.

Step-by-step explanation:

As given the equation in the form

\frac{1}{x}-\frac{1}{x-2} = 3

x

1

−

x−2

1

=3

Simplify the above equation

(x-2)-x = 3x × (x-2)

x-2 - x = 3x² - 6x

3x² - 6x + 2 = 0

As the equation is written in the form ax² + bx + c = 0

x =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}x=

2a

−b±

b

2

−4ac

a = 3 , b = -6 , c = 2

Put all the values in the above equation

x =\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times 3\times 2}}{2\times 3}x=

2×3

−(−6)±

(−6)

2

−4×3×2

x =\frac{6\pm\sqrt{36-24}}{6}x=

6

6±

36−24

x =\frac{6\pm\sqrt{12}}{6}x=

6

6±

12

x =\frac{6\pm2\sqrt{3}}{6}x=

6

6±2

3

x =\frac{3\pm1\sqrt{3}}{3}x=

3

3±1

3

Thus

x =\frac{3+1\sqrt{3}}{3}x=

3

3+1

3

x =\frac{3-1\sqrt{3}}{3}x=

3

3−1

3

Therefore the roots of the quadratic equation \frac{1}{x}-\frac{1}{x-2} = 3

x

1

−

x−2

1

=3 are x =\frac{3+1\sqrt{3}}{3}x=

3

3+1

3

,x =\frac{3-1\sqrt{3}}{3}x=

3

3−1

3

.