Solve Даваи этот реши (3x - x ^ 2)/2 + (2x ^ 2 - x)/6 = x | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac { 3 x - x ^ { 2 } } { 2 } + \frac { 2 x ^ { 2 } - x } { 6 } = x$$

Solve for x

$x=2$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2$ and $6$ is $6$. Multiply $\frac{3x-x^{2}}{2}$ times $\frac{3}{3}$.

$$\frac{3\left(3x-x^{2}\right)}{6}+\frac{2x^{2}-x}{6}=x$$

Since $\frac{3\left(3x-x^{2}\right)}{6}$ and $\frac{2x^{2}-x}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{3\left(3x-x^{2}\right)+2x^{2}-x}{6}=x$$

Do the multiplications in $3\left(3x-x^{2}\right)+2x^{2}-x$.

$$\frac{9x-3x^{2}+2x^{2}-x}{6}=x$$

Combine like terms in $9x-3x^{2}+2x^{2}-x$.

$$\frac{8x-x^{2}}{6}=x$$

Divide each term of $8x-x^{2}$ by $6$ to get $\frac{4}{3}x-\frac{1}{6}x^{2}$.

$$\frac{4}{3}x-\frac{1}{6}x^{2}=x$$

Subtract $x$ from both sides.

$$\frac{4}{3}x-\frac{1}{6}x^{2}-x=0$$

Combine $\frac{4}{3}x$ and $-x$ to get $\frac{1}{3}x$.

$$\frac{1}{3}x-\frac{1}{6}x^{2}=0$$

Factor out $x$.

$$x\left(\frac{1}{3}-\frac{1}{6}x\right)=0$$

To find equation solutions, solve $x=0$ and $\frac{1}{3}-\frac{x}{6}=0$.

$$x=0$$ $$x=2$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2$ and $6$ is $6$. Multiply $\frac{3x-x^{2}}{2}$ times $\frac{3}{3}$.

$$\frac{3\left(3x-x^{2}\right)}{6}+\frac{2x^{2}-x}{6}=x$$

Since $\frac{3\left(3x-x^{2}\right)}{6}$ and $\frac{2x^{2}-x}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{3\left(3x-x^{2}\right)+2x^{2}-x}{6}=x$$

Do the multiplications in $3\left(3x-x^{2}\right)+2x^{2}-x$.

$$\frac{9x-3x^{2}+2x^{2}-x}{6}=x$$

Combine like terms in $9x-3x^{2}+2x^{2}-x$.

$$\frac{8x-x^{2}}{6}=x$$

Divide each term of $8x-x^{2}$ by $6$ to get $\frac{4}{3}x-\frac{1}{6}x^{2}$.

$$\frac{4}{3}x-\frac{1}{6}x^{2}=x$$

Subtract $x$ from both sides.

$$\frac{4}{3}x-\frac{1}{6}x^{2}-x=0$$

Combine $\frac{4}{3}x$ and $-x$ to get $\frac{1}{3}x$.

$$\frac{1}{3}x-\frac{1}{6}x^{2}=0$$

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.

$$-\frac{1}{6}x^{2}+\frac{1}{3}x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-\frac{1}{6}$ for $a$, $\frac{1}{3}$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\frac{1}{3}±\sqrt{\left(\frac{1}{3}\right)^{2}}}{2\left(-\frac{1}{6}\right)}$$

Take the square root of $\left(\frac{1}{3}\right)^{2}$.

$$x=\frac{-\frac{1}{3}±\frac{1}{3}}{2\left(-\frac{1}{6}\right)}$$

Multiply $2$ times $-\frac{1}{6}$.

$$x=\frac{-\frac{1}{3}±\frac{1}{3}}{-\frac{1}{3}}$$

Now solve the equation $x=\frac{-\frac{1}{3}±\frac{1}{3}}{-\frac{1}{3}}$ when $±$ is plus. Add $-\frac{1}{3}$ to $\frac{1}{3}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{-\frac{1}{3}}$$

Divide $0$ by $-\frac{1}{3}$ by multiplying $0$ by the reciprocal of $-\frac{1}{3}$.

$$x=0$$

Now solve the equation $x=\frac{-\frac{1}{3}±\frac{1}{3}}{-\frac{1}{3}}$ when $±$ is minus. Subtract $\frac{1}{3}$ from $-\frac{1}{3}$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide $-\frac{2}{3}$ by $-\frac{1}{3}$ by multiplying $-\frac{2}{3}$ by the reciprocal of $-\frac{1}{3}$.

$$x=2$$

The equation is now solved.

$$x=0$$ $$x=2$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2$ and $6$ is $6$. Multiply $\frac{3x-x^{2}}{2}$ times $\frac{3}{3}$.

$$\frac{3\left(3x-x^{2}\right)}{6}+\frac{2x^{2}-x}{6}=x$$

Since $\frac{3\left(3x-x^{2}\right)}{6}$ and $\frac{2x^{2}-x}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{3\left(3x-x^{2}\right)+2x^{2}-x}{6}=x$$

Do the multiplications in $3\left(3x-x^{2}\right)+2x^{2}-x$.

$$\frac{9x-3x^{2}+2x^{2}-x}{6}=x$$

Combine like terms in $9x-3x^{2}+2x^{2}-x$.

$$\frac{8x-x^{2}}{6}=x$$

Divide each term of $8x-x^{2}$ by $6$ to get $\frac{4}{3}x-\frac{1}{6}x^{2}$.

$$\frac{4}{3}x-\frac{1}{6}x^{2}=x$$

Subtract $x$ from both sides.

$$\frac{4}{3}x-\frac{1}{6}x^{2}-x=0$$

Combine $\frac{4}{3}x$ and $-x$ to get $\frac{1}{3}x$.

$$\frac{1}{3}x-\frac{1}{6}x^{2}=0$$

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 $x^{2}+bx=c$.

$$-\frac{1}{6}x^{2}+\frac{1}{3}x=0$$

Multiply both sides by $-6$.

$$\frac{-\frac{1}{6}x^{2}+\frac{1}{3}x}{-\frac{1}{6}}=\frac{0}{-\frac{1}{6}}$$

Dividing by $-\frac{1}{6}$ undoes the multiplication by $-\frac{1}{6}$.

$$x^{2}+\frac{\frac{1}{3}}{-\frac{1}{6}}x=\frac{0}{-\frac{1}{6}}$$

Divide $\frac{1}{3}$ by $-\frac{1}{6}$ by multiplying $\frac{1}{3}$ by the reciprocal of $-\frac{1}{6}$.

$$x^{2}-2x=\frac{0}{-\frac{1}{6}}$$

Divide $0$ by $-\frac{1}{6}$ by multiplying $0$ by the reciprocal of $-\frac{1}{6}$.

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

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

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=1$$ $$x-1=-1$$

Add $1$ to both sides of the equation.

$$x=2$$ $$x=0$$