Solve 5x + 2 = 2 - x ^ 2 ) единствен ) ровно пят | 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

$$5 x + 2 = 2 - 2 x ^ { 2 }$$

Solve for x

$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Subtract $2$ from both sides.

$$5x+2-2=-2x^{2}$$

Subtract $2$ from $2$ to get $0$.

$$5x=-2x^{2}$$

Add $2x^{2}$ to both sides.

$$5x+2x^{2}=0$$

Factor out $x$.

$$x\left(5+2x\right)=0$$

To find equation solutions, solve $x=0$ and $5+2x=0$.

$$x=0$$ $$x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract $2$ from both sides.

$$5x+2-2=-2x^{2}$$

Subtract $2$ from $2$ to get $0$.

$$5x=-2x^{2}$$

Add $2x^{2}$ to both sides.

$$5x+2x^{2}=0$$

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.

$$2x^{2}+5x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $5$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 2}$$

Take the square root of $5^{2}$.

$$x=\frac{-5±5}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-5±5}{4}$$

Now solve the equation $x=\frac{-5±5}{4}$ when $±$ is plus. Add $-5$ to $5$.

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

Divide $0$ by $4$.

$$x=0$$

Now solve the equation $x=\frac{-5±5}{4}$ when $±$ is minus. Subtract $5$ from $-5$.

$$x=-\frac{10}{4}$$

Reduce the fraction $\frac{-10}{4}$ to lowest terms by extracting and canceling out $2$.

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

The equation is now solved.

$$x=0$$ $$x=-\frac{5}{2}$$

Steps for Completing the Square

Add $2x^{2}$ to both sides.

$$5x+2+2x^{2}=2$$

Subtract $2$ from both sides.

$$5x+2x^{2}=2-2$$

Subtract $2$ from $2$ to get $0$.

$$5x+2x^{2}=0$$

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

$$2x^{2}+5x=0$$

Divide both sides by $2$.

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

Dividing by $2$ undoes the multiplication by $2$.

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

Divide $0$ by $2$.

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

Divide $\frac{5}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{4}$. Then add the square of $\frac{5}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $\frac{5}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor $x^{2}+\frac{5}{2}x+\frac{25}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{5}{4}$$ $$x+\frac{5}{4}=-\frac{5}{4}$$

Subtract $\frac{5}{4}$ from both sides of the equation.

$$x=0$$ $$x=-\frac{5}{2}$$