Enter an equation or problem

Camera input is not recognized!

Solution - Solving quadratic equations by completing the square

Exact form:

x_{1} = - 1 + \frac{1 \cdot \sqrt{3}}{3}

x_1=-1+\frac{1\cdot \sqrt{3}}{3}

x_{2} = - 1 - \frac{1 \cdot \sqrt{3}}{3}

x_2=-1-\frac{1\cdot \sqrt{3}}{3}

Decimal form:

x_{1} = - 0.423

x_1=-0.423

x_{2} = - 1.577

x_2=-1.577

See steps

Step-by-step explanation

1. Identify the coefficients

Use the standard form of a quadratic equation, $a x^{2} + b x + c = 0$ , to find the coefficients:

$3 x^{2} + 6 x + 2 = 0$

$a = 3$
$b = 6$
$c = 2$

2. Make the a coefficient equal 1

Because $a = 3$ , divide all coefficients and constants on both sides of the equation by $3$ :

$3 x^{2} + 6 x + 2 = 0$

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

Simplify the expression

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

The coefficients are:
$a = 1$
$b = 2$
$c = \frac{2}{3}$

3. Move the constant to the right side of the equation and combine

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

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

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

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

4. Complete the square

To make the left side of the equation into a perfect square trinomial, add a new constant equal to ${(\frac{b}{2})}^{2}$ to the equation:

$b = 2$

${(\frac{b}{2})}^{2} = {(\frac{2}{2})}^{2}$

Use the exponents fraction rule ${(\frac{x}{y})}^{2} = \frac{x^{2}}{y^{2}}$

${(\frac{2}{2})}^{2} = \frac{2^{2}}{2^{2}}$

$\frac{2^{2}}{2^{2}} = \frac{4}{4}$

$\frac{4}{4} = 1$

Add $1$ to both sides of the equation:

3 additional steps

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

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Now we have perfect square trinomial, we can write it as a perfect square form by adding half of the $b$ coefficient, $\frac{b}{2}$ :
$b = 2$

2 additional steps

$\frac{b}{2} = \frac{2}{2}$

Find the greatest common factor of the numerator and denominator:

$\frac{b}{2} = \frac{(1 \cdot 2)}{(1 \cdot 2)}$

Factor out and cancel the greatest common factor:

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

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

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

5. Solve for $x$

Take the square root of both sides of the equation: IMPORTANT: When finding the square root of a constant, we get two solutions: positive and negative

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

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

Cancel out the square and square root on the left side of the equation:

$x + 1 = \pm \sqrt{\frac{1}{3}}$

Subtract $1$ from both sides

$x + 1 - 1 = - 1 \pm \sqrt{\frac{1}{3}}$

Simplify the left side:

$x = - 1 \pm \sqrt{\frac{1}{3}}$

$x = - 1 \pm \frac{\sqrt{1}}{\sqrt{3}}$

$x = - 1 \pm \frac{1}{\sqrt{3}}$

$x = - 1 \pm \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$

$x = - 1 \pm \frac{1 \cdot \sqrt{3}}{3}$

$x_{1} = - 1 + \frac{1 \cdot \sqrt{3}}{3}$
$x_{2} = - 1 - \frac{1 \cdot \sqrt{3}}{3}$

How did we do?

Please leave us feedback.

Why learn this

In their most basic function, quadratic equations define shapes like circles, ellipses and parabolas. These shapes can in turn be used to predict the curve of an object in motion, such as a ball kicked by football player or shot out of a cannon.
When it comes to an object’s movement through space, what better place to start than space itself, with the revolution of planets around the sun in our solar system. The quadratic equation was used to establish that planets’ orbits are elliptical, not circular. Determining the path and speed an object travels through space is possible even after it has come to a stop: the quadratic equation can calculate how fast a vehicle was moving when it crashed. With information like this, the automotive industry can design brakes to prevent collisions in the future. Many industries use the quadratic equation to predict and thus improve their products’ lifespan and safety.

Terms and topics

Solving quadratic equations by completing the square