Algebra Examples

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

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

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 2 \cdot - 1 = - 2

$a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is

b = - 1

$b = - 1$ .

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Step 1.1.1

Factor

- 1

$- 1$ out of

- x

$- x$ .

2 x^{2} - (x) - 1 = 0

$2 x^{2} - (x) - 1 = 0$

Step 1.1.2

Rewrite

- 1

$- 1$ as

1

$1$ plus

- 2

$- 2$

2 x^{2} + (1 - 2) x - 1 = 0

$2 x^{2} + (1 - 2) x - 1 = 0$

Step 1.1.3

Apply the distributive property.

2 x^{2} + 1 x - 2 x - 1 = 0

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

Step 1.1.4

Multiply

x

$x$ by

1

$1$ .

2 x^{2} + x - 2 x - 1 = 0

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

2 x^{2} + x - 2 x - 1 = 0

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

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

(2 x^{2} + x) - 2 x - 1 = 0

$(2 x^{2} + x) - 2 x - 1 = 0$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

x (2 x + 1) - (2 x + 1) = 0

$x (2 x + 1) - (2 x + 1) = 0$

x (2 x + 1) - (2 x + 1) = 0

$x (2 x + 1) - (2 x + 1) = 0$

Step 1.3

Factor the polynomial by factoring out the greatest common factor,

2 x + 1

$2 x + 1$ .

(2 x + 1) (x - 1) = 0

$(2 x + 1) (x - 1) = 0$

(2 x + 1) (x - 1) = 0

$(2 x + 1) (x - 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

2 x + 1 = 0

$2 x + 1 = 0$

x - 1 = 0

$x - 1 = 0$

Step 3

Set

2 x + 1

$2 x + 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

2 x + 1

$2 x + 1$ equal to

0

$0$ .

2 x + 1 = 0

$2 x + 1 = 0$

Step 3.2

Solve

2 x + 1 = 0

$2 x + 1 = 0$ for

x

$x$ .

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Step 3.2.1

Subtract

1

$1$ from both sides of the equation.

2 x = - 1

$2 x = - 1$

Step 3.2.2

Divide each term in

2 x = - 1

$2 x = - 1$ by

2

$2$ and simplify.

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Step 3.2.2.1

Divide each term in

2 x = - 1

$2 x = - 1$ by

2

$2$ .

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

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

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

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

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

Step 3.2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{- 1}{2}

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

x = \frac{- 1}{2}

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

x = \frac{- 1}{2}

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

Step 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Move the negative in front of the fraction.

x = - \frac{1}{2}

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

x = - \frac{1}{2}

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

x = - \frac{1}{2}

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

x = - \frac{1}{2}

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

x = - \frac{1}{2}

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

Step 4

Set

x - 1

$x - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

x - 1

$x - 1$ equal to

0

$0$ .

x - 1 = 0

$x - 1 = 0$

Step 4.2

Add

1

$1$ to both sides of the equation.

x = 1

$x = 1$

x = 1

$x = 1$

Step 5

The final solution is all the values that make

(2 x + 1) (x - 1) = 0

$(2 x + 1) (x - 1) = 0$ true.

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

$x = - \frac{1}{2}, 1$