Algebra Examples

$\frac{4}{5} x^{2} = 2 x - \frac{4}{5}$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $2 x$ from both sides of the equation.

$\frac{4}{5} x^{2} - 2 x = - \frac{4}{5}$

Step 1.2

Add $\frac{4}{5}$ to both sides of the equation.

$\frac{4}{5} x^{2} - 2 x + \frac{4}{5} = 0$

Step 2

Combine $\frac{4}{5}$ and $x^{2}$ .

$\frac{4 x^{2}}{5} - 2 x + \frac{4}{5} = 0$

Step 3

Factor $2$ out of $\frac{4 x^{2}}{5} - 2 x + \frac{4}{5}$ .

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Step 3.1

Factor $2$ out of $\frac{4 x^{2}}{5}$ .

$2 (\frac{2 x^{2}}{5}) - 2 x + \frac{4}{5} = 0$

Step 3.2

Factor $2$ out of $- 2 x$ .

$2 (\frac{2 x^{2}}{5}) + 2 (- x) + \frac{4}{5} = 0$

Step 3.3

Factor $2$ out of $\frac{4}{5}$ .

$2 (\frac{2 x^{2}}{5}) + 2 (- x) + 2 (\frac{2}{5}) = 0$

Step 3.4

Factor $2$ out of $2 \frac{2 x^{2}}{5} + 2 (- x)$ .

$2 (\frac{2 x^{2}}{5} - x) + 2 (\frac{2}{5}) = 0$

Step 3.5

Factor $2$ out of $2 (\frac{2 x^{2}}{5} - x) + 2 (\frac{2}{5})$ .

$2 (\frac{2 x^{2}}{5} - x + \frac{2}{5}) = 0$

Step 4

To write $- x$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$2 (\frac{2 x^{2}}{5} - x \cdot \frac{5}{5} + \frac{2}{5}) = 0$

Step 5

Combine $- x$ and $\frac{5}{5}$ .

$2 (\frac{2 x^{2}}{5} + \frac{- x \cdot 5}{5} + \frac{2}{5}) = 0$

Step 6

Combine the numerators over the common denominator.

$2 (\frac{2 x^{2} - x \cdot 5}{5} + \frac{2}{5}) = 0$

Step 7

Combine the numerators over the common denominator.

$2 (\frac{2 x^{2} - x \cdot 5 + 2}{5}) = 0$

Step 8

Multiply $5$ by $- 1$ .

$2 (\frac{2 x^{2} - 5 x + 2}{5}) = 0$

Step 9

Factor by grouping.

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Step 9.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

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Step 9.1.1

Factor $- 5$ out of $- 5 x$ .

$2 (\frac{2 x^{2} - 5 x + 2}{5}) = 0$

Step 9.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$2 (\frac{2 x^{2} + (- 1 - 4) x + 2}{5}) = 0$

Step 9.1.3

Apply the distributive property.

$2 (\frac{2 x^{2} - 1 x - 4 x + 2}{5}) = 0$

Step 9.2

Factor out the greatest common factor from each group.

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Step 9.2.1

Group the first two terms and the last two terms.

$2 (\frac{(2 x^{2} - 1 x) - 4 x + 2}{5}) = 0$

Step 9.2.2

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

$2 (\frac{x (2 x - 1) - 2 (2 x - 1)}{5}) = 0$

Step 9.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$2 (\frac{(2 x - 1) (x - 2)}{5}) = 0$

Step 10

Combine $2$ and $\frac{(2 x - 1) (x - 2)}{5}$ .

$\frac{2 (2 x - 1) (x - 2)}{5} = 0$

Step 11

Set the numerator equal to zero.

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

Step 12

Solve the equation for $x$ .

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Step 12.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 1 = 0$

$x - 2 = 0$

Step 12.2

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 12.2.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 12.2.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 12.2.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 12.2.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 12.2.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 12.2.2.2.2

Simplify the left side.

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Step 12.2.2.2.2.1

Cancel the common factor of $2$ .

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Step 12.2.2.2.2.1.1

Cancel the common factor.

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

Step 12.2.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 12.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 12.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 12.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 12.4

The final solution is all the values that make $2 (2 x - 1) (x - 2) = 0$ true.

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