Solve (x - 1/2) ^ 2 + 1/2 * (x - 1) ^ 2 = 3/4 | 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

$${(x-\frac{1}{2})}^{2}+\frac{1}{2}(x-1)^{2}=\frac{3}{4}$$

Solve for x

$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-\frac{1}{2}\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x-1\right)^{2}=\frac{3}{4}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x^{2}-2x+1\right)=\frac{3}{4}$$

Use the distributive property to multiply $\frac{1}{2}$ by $x^{2}-2x+1$.

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

Combine $x^{2}$ and $\frac{1}{2}x^{2}$ to get $\frac{3}{2}x^{2}$.

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

Combine $-x$ and $-x$ to get $-2x$.

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

Add $\frac{1}{4}$ and $\frac{1}{2}$ to get $\frac{3}{4}$.

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

Subtract $\frac{3}{4}$ from both sides.

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

Subtract $\frac{3}{4}$ from $\frac{3}{4}$ to get $0$.

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

Factor out $x$.

$$x\left(\frac{3}{2}x-2\right)=0$$

To find equation solutions, solve $x=0$ and $\frac{3x}{2}-2=0$.

$$x=0$$ $$x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-\frac{1}{2}\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x-1\right)^{2}=\frac{3}{4}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x^{2}-2x+1\right)=\frac{3}{4}$$

Use the distributive property to multiply $\frac{1}{2}$ by $x^{2}-2x+1$.

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

Combine $x^{2}$ and $\frac{1}{2}x^{2}$ to get $\frac{3}{2}x^{2}$.

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

Combine $-x$ and $-x$ to get $-2x$.

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

Add $\frac{1}{4}$ and $\frac{1}{2}$ to get $\frac{3}{4}$.

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

Subtract $\frac{3}{4}$ from both sides.

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

Subtract $\frac{3}{4}$ from $\frac{3}{4}$ to get $0$.

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

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

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

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

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

The opposite of $-2$ is $2$.

$$x=\frac{2±2}{2\times \frac{3}{2}}$$

Multiply $2$ times $\frac{3}{2}$.

$$x=\frac{2±2}{3}$$

Now solve the equation $x=\frac{2±2}{3}$ when $±$ is plus. Add $2$ to $2$.

$$x=\frac{4}{3}$$

Now solve the equation $x=\frac{2±2}{3}$ when $±$ is minus. Subtract $2$ from $2$.

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

Divide $0$ by $3$.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{3}$$ $$x=0$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-\frac{1}{2}\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x-1\right)^{2}=\frac{3}{4}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$x^{2}-x+\frac{1}{4}+\frac{1}{2}\left(x^{2}-2x+1\right)=\frac{3}{4}$$

Use the distributive property to multiply $\frac{1}{2}$ by $x^{2}-2x+1$.

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

Combine $x^{2}$ and $\frac{1}{2}x^{2}$ to get $\frac{3}{2}x^{2}$.

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

Combine $-x$ and $-x$ to get $-2x$.

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

Add $\frac{1}{4}$ and $\frac{1}{2}$ to get $\frac{3}{4}$.

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

Subtract $\frac{3}{4}$ from both sides.

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

Subtract $\frac{3}{4}$ from $\frac{3}{4}$ to get $0$.

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

Divide both sides of the equation by $\frac{3}{2}$, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by $\frac{3}{2}$ undoes the multiplication by $\frac{3}{2}$.

$$x^{2}+\left(-\frac{2}{\frac{3}{2}}\right)x=\frac{0}{\frac{3}{2}}$$

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

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

Divide $0$ by $\frac{3}{2}$ by multiplying $0$ by the reciprocal of $\frac{3}{2}$.

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

Divide $-\frac{4}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{2}{3}$. Then add the square of $-\frac{2}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\left(-\frac{2}{3}\right)^{2}$$

Square $-\frac{2}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Factor $x^{2}-\frac{4}{3}x+\frac{4}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add $\frac{2}{3}$ to both sides of the equation.

$$x=\frac{4}{3}$$ $$x=0$$