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**problem**

Solve $\color{blue}{2x^2+8x-10 = 0}$ by completing the square.

**solution**

**The solutions are:**

**explanation**

**Step 1:** Divide equation by a number in front of the squared term. In this case we will divide by $ 2 $.

**Step 2:** Keep all terms containing $ x $ on one side. Move $ -5 $ to the right.

**Step 3:** Take half of the x -term coefficient and square it. Add this value to both sides.

The x-term coefficient = $ 4 $

The half of the x-term coefficient = $ 2 $

After squaring we have $ 2^2 = 4 $

When we add $ 4 $ to both sides we have:

$$ x^2+4x+4 = 5 + 4 $$**Step 4:** Simplify right side.

**Step 5:** Write the perfect square on the left.

**Step 6:** Take the square root of both sides.

**Step 7:** Solve for $ x $.

$ x_1,x_2 = - 2 \pm \sqrt{ 9 } $

that is,

$ x_1 = -5 $

$ x_2 = 1 $

working...

EXAMPLES

example 1:ex 1:

Solve for $x^2 + 3x - 4 = 0$ by factoring.

example 2:ex 2:

Solve $4x^2 - x - 3 = 0$ by completing the square.

example 3:ex 3:

Solve $-2x^2 - 0.5x + 0.75 = 0$ using the quadratic formula.

example 4:ex 4:

Solve $ \frac{2}{3} x^2 - \frac{1}{3} x - 5 = 0 $.

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TUTORIAL

The most commonly used methods for solving quadratic equations are:

**1**. Factoring method

**2**. 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 $ x^2 \color{red}{-8}x \color{blue}{+ 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 is $ a + b = \color{red}{8} $ and a product of $ a \cdot b = \color{blue}{15} $.

After some trials and errors, we see that $ a = 3 $ and $ b = 5 $.

Now we use formula $ x^2 - 8x + 15 = (x - a)(x - b) $ to get factored form:

$$ x^2 - 8x + 15 = (x - 3)(x - 5) $$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 $ x^2 -8x = 0 $ by factoring.

In this case, (when the coefficient **c = 0** ) we can factor out $ \color{blue}{x} $ out of $ x^2 - 8x $.

**Example 03:** Solve $ x^2 - 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 $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)

**Step 2**:Plug the values for a, b, and c into the quadratic formula and simplify.

**Step 3**: Solve for $x_1$ and $x_2$

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 $ 2x^2 + 8x - 10= 0$ by completing the square.

**Step 1**: Divide the equation by the number in front of the square term.

**Step 2**: move $-5$ to the right:

**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.

**Step 4**: Simplify left and right side.

**Step 5**: Write the perfect square on the left.

**Step 6**: Take the square root of both sides.

**Step 7**: Solve for $x_1$ and $x_2$ .

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