This calculator solves quadratic equations using three different methods : the quadratic formula method, completing the square, and the factoring method. Calculator shows all the work and provides detailed explanation on how to solve an equation.
Solve $\color{blue}{\dfrac{2}{3}x^2 = 0}$ using factoring.
solution
$$ \color{blue}{ x = 0 }$$explanation
First we need to factor trinomial $ \color{blue}{ \dfrac{2}{3}x^2 } $ and than we use factored form to solve an equation $ \color{blue}{ \dfrac{2}{3}x^2 = 0} $.
Step 1: We can simplify equation by multiplying both sides by 3. After multiplying we have the following equation:
$$ \begin{aligned} \dfrac{2}{3}x^2 &= 0 \,\,\, / \color{orangered}{\cdot \, 3 } \\[0.9 em ] 2x^2 &=0 \end{aligned} $$Step 2: Simplify equation by dividing all coefficients by 2
$$ \begin{aligned} 2x^2 &= 0 \,\,\, / \color{orangered}{ : 2 } \\[0.9 em ] x^2 &=0 \end{aligned} $$Since the quadratic trinomial cannot be factored out we solved the problem using quadratic formula.
The most commonly used methods for solving quadratic equations are:
1. Factoring method
2. Solving quadratic equations by completing the square
3. Using quadratic formula
In the following sections, we'll go over these methods.
If a quadratic trinomial can be factored, this is the best solving method.
We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.
Example 01: Solve x2-8x+15=0 by factoring.
Here we see that the leading coefficient is 1, so the factoring method is our first choice.
To factor this equation, we must find two numbers a and b with a sum of a + b = 8 and a product of a × b = 15
After some trials and errors, we see that a = 3 and b = 5.
Now we use a formula x2-8x+15=(x-a)(x-b) to get factored form:
x2-8x+15=(x-3)(x-5)
Divide the factored form into two linear equations to get solutions.
$$ \begin{aligned} x^2 - 8x + 15 &= 0 \\ (x - 3)(x - 5) &= 0 \\ x -3 &= 0 ~~ \text{or} ~~ x - 5= 0 \\ x &= 3 ~~ \text{or} ~~ x = 5 \end{aligned} $$Example 02: Solve x2-8x=0 by factoring.
In this case, (when the coefficient c = 0) we can factor out x out of x2-8.
$$ \begin{aligned} x^2 - 8x &= 0 \\ \color{blue}{x} \cdot ( x - 8 ) &= 0 \\ x &= 0 ~~ \text{or} ~~ x - 8 = 0 \\ x &= 0 ~~ \text{or} ~~ x = 8 \end{aligned} $$Example 03: Solve x2-16=0 by factoring.
In this case, (when the middle term is equal 0) we can use the difference of squares formula.
$$ \begin{aligned} x^2 - 16 &= 0 \\ x^2 - 4^2 &= 0 \text{ use } a^2 - b^2 = (a-b)(a+b) \\ (x - 4)(x+4) &= 0 \\ x - 4 &= 0 ~~ \text{or} ~~ x + 4 = 0 \\ x &= 4 ~~ \text{or} ~~ x = -4 \end{aligned} $$This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.
Example 05: Solve equation $ 2x^2 + 3x - 2 = 0$ by using quadratic formula.
Step 1: Read the values of $a$, $b$, and $c$ from the quadratic equation. (a is the number in front of x2, b is the number in front of x and c is the number at the end)
a = 2, b = 3 and c = -2
Step 2:Plug the values for a, b, and c into the quadratic formula and simplify.
$$ \begin{aligned} x_1, x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-2) }}{2 \cdot 2} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{9+ 16 }}{4} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{25}}{4} \\ x_1, x_2 &= \frac{-3 \pm 5}{4} \end{aligned} $$Step 3: Solve for x1 and x2
$$ \begin{aligned} x_1 = & \frac{-3 \color{blue}{+} 5}{4} = \frac{2}{4} = \frac{1}{2} \\ x_2 = & \frac{-3 \color{blue}{-} 5}{4} = \frac{-8}{4} = -2 \end{aligned} $$This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.
Example 04: Solve equation 2x2+8x-10=0 by completing the square.
Step 1: Divide the equation by the number in front of the square term.
$$ \begin{aligned} 2x^2 + 8x - 10 & = 0 ~~ / ~ \color{orangered}{:2} \\ \frac{2x^2}{2} + \frac{8x}{2} - \frac{10}{2} & = \frac{0}{2} \\ x^2 + 4x - 5 & = 0 \end{aligned} $$Step 2: move -5 to the right:
x2+4x=5
Step 3: Take half of the x-term coefficient $ \color{blue}{\dfrac{4}{2}} $, square it $ \color{blue}{\left(\dfrac{4}{2} \right)^2} $ and add this value to both sides.
$$ x^2 + 4x + \color{blue}{\left(\frac{4}{2} \right)^2} = 5 + \color{blue}{\left(\frac{4}{2} \right)^2} $$Step 4: Simplify left and right side.
x2+4x+22=9
Step 5: Write the perfect square on the left.
$$ \left( x + 2 \right)^2 = 9 $$Step 6: Take the square root of both sides.
$$ \begin{aligned} x + 2 &= \pm \sqrt{9} \\\\ x + 2 &= \pm 3 \end{aligned} $$Step 7: Solve for $x_1$ and $x_2$ .
$$ \begin{aligned} x_1 & = +3 - 2 = 1 \\ x_2 & = -3 - 2 = - 5 \end{aligned} $$1. Quadratic Equation — step-by-step examples, video tutorials with worked examples.
2. Completing the Square — video on Khan Academy
3. Completing the Square — video on Khan Academy