Solve x ^ 2 - (- 1 + (- 1)) * x + (- x - 1) = 0 | 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

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

Solve for x

$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$
$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $1$ from $-1$ to get $-2$.

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

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

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

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

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

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

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

Square $1$.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)}}{2}$$

Multiply $-4$ times $-1$.

$$x=\frac{-1±\sqrt{1+4}}{2}$$

Add $1$ to $4$.

$$x=\frac{-1±\sqrt{5}}{2}$$

Now solve the equation $x=\frac{-1±\sqrt{5}}{2}$ when $±$ is plus. Add $-1$ to $\sqrt{5}$.

$$x=\frac{\sqrt{5}-1}{2}$$

Now solve the equation $x=\frac{-1±\sqrt{5}}{2}$ when $±$ is minus. Subtract $\sqrt{5}$ from $-1$.

$$x=\frac{-\sqrt{5}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-1}{2}$$ $$x=\frac{-\sqrt{5}-1}{2}$$

Steps for Completing the Square

Subtract $1$ from $-1$ to get $-2$.

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

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

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

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

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

Add $1$ to both sides. Anything plus zero gives itself.

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

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

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

Square $\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

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

Add $1$ to $\frac{1}{4}$.

$$x^{2}+x+\frac{1}{4}=\frac{5}{4}$$

Factor $x^{2}+x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{5}}{2}$$ $$x+\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Subtract $\frac{1}{2}$ from both sides of the equation.

$$x=\frac{\sqrt{5}-1}{2}$$ $$x=\frac{-\sqrt{5}-1}{2}$$