QUADRATIC EQUATION BY COMPLETING THE SQUARE

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Generally, a quadratic is usually in the form:
$$Ax^2+bx+c=0$$

However, solving a quadratic equation using the completing the square method requires that the coefficient of $x^2$ be a perfect square.

Take $4x^2+15x+6$, for example. This can be solved using completing the square because the coefficient of $x^2$ is 4 which  is a perfect square because

 $$4=2^2$$

But, $2x^2+15x+6$ can not be solved using completing the square because the coefficient of $x^2$ is 2( which is not a perfect square).

"what if the former is the case?"

Don't worry, you just divide through by the coefficient of $x^2$ like this 

$\frac{2x^2}{2}+\frac{15x}{2}+\frac{6}{2}=0$

${x^2}+\frac{15x}{2}+\frac{6}{2}=0$

As you can see, the coefficient of $x^2$, which is 1, is a perfect square. 

👉This post is an extension of our post on quadratic equation👈

Today, we'll look at four examples, but first, let's illustrate quadratic equations by completing the square method

Suppose:
$$x^2+8x+7=0$$

The first step is to rearrange the equation so that the $x^2$ and $x$ coefficients are on one side of the equation, and the constant is on the other. 

$$x^2+8x=-7$$

What next?

Add the square half of the coefficient of $x$ to both sides.

In this case, the coefficient of $x$ is 8, Hence the square half will be $(\frac{8}{2})^2=16$. 

Accordingly,

$$x^2+8x+16=-7+16$$

Now, let solve the simple equation at the right-hand side of the equation

$$x^2+8x+16=9$$

Wait! did you notice that I didn't solve it together? Yes, we have a complete 
square on the left-hand side.

This means $x^2+8x+16$ is a square of $(x+4)$, Hence

$$(x+4)^2=9$$

Next, we add the square root to both sides

$$\sqrt{(x+4)^2}=\sqrt{9}$$

$$x+4=\pm 3$$

$$x=\pm 3-4$$

Now let's split the $\pm$

$$x=+3-4,-3-4$$

$$x=-1, -7$$

Let's plunge in:

Example 1

Solve $x^2+4x-16=0$ using completing the square.

Solution

First, let's re-arrange
$x^2+4x=16$

Add the square half of the coefficient of $x$ to both sides.

$x^2+4x+(\frac{4}{2})^2=16+(\frac{4}{2})^2$

Recalled that $\frac{4}{2}=2$, hence,

$x^2+4x+(2)^2=16+(2)^2$

Adding the right-hand side

$x^2+4x+(2)^2=16+4$

$x^2+4x+(2)^2=20$

$x^2+4x+4=20$

The left-hand side is a perfect square$(x+2)^2=x^2+4x+4$, hence
$(x+2)^2=20$ 

Taking the square root of both sides
$\sqrt{(x+2)^2}=\pm\sqrt{20}$
$x+2=\pm\sqrt{20}$

Note, the square root of a number will give a $\pm$ answer

$x+2=\pm4.47$ 

$x=\pm4.47-2$

$x=+4.47-2$ or $-4.47-2$

$x=2.27$ or $-6.47$

Example 2

Using competing for the square, solve the quadratic equation: $x^2-14x-15=0$

Solution

First, you re-arrange the equation
$x^2-14x=15$

Adding the square half the coefficient of $x$ to both sides.

$x^2-14x+(\frac{-14}{2})^2=15+(\frac{-14}{2})^2$

Recalled $(\frac{-14}{2})=-7$, Hence
$x^2-14x+(-7)^2=15+(-7)^2$

Solving the right-hand side
$x^2-14x+(-7)^2=15+49$

$x^2-14x+49=64$

The left-hand side is a perfect square because $(x-7)^2=x^2-14x+49$. Thus
$(x-7)^2=64$

Adding square root to both sides

$\sqrt{(x-7)^2}=\sqrt{64}$

$x-7=\pm 8$

$x=\pm 8+7$

$x=8+7$ or $-8+7$

$x=15$ or $-1$

Example 3

Solve $5x^2+7x-6=0$ using completing the square method.

Solution

This question is slightly different from the two previous questions because the coefficient of $x^2$(5) is not a perfect square. 

Remember that solving with completing the square method requires that the coefficient of $x^2$ be a perfect square.

So, what should we do?

We divide both through by 5 so that the coefficient of $x^2$ will equal 1 (which is a perfect square)

$\frac{\require{cancel}\bcancel{5}x^2}{\require{cancel}\bcancel{5}}+\frac{7x}{5}-\frac{-6}{5}=\frac{0}{5}$

$x^2+\frac{7x}{5}-\frac{-6}{5}=0$

As usual, let's re-arrange the equation
$x^2+\frac{7x}{5}=\frac{6}{5}$

Add the square half of the coefficient of $x $ to both sides;

Note: Half of the coefficient is:
$\frac{7}{5}÷2$

This translates to
$\frac{7}{10}$, this square half is ($\frac{7}{10})^2$

Accordingly,
$x^2+\frac{7x}{5}+(\frac{7}{10})^2=\frac{6}{5}+(\frac{7}{10})^2$

Adding the right-hand side

$x^2+\frac{7x}{5}+(\frac{7}{10})^2=\frac{6}{5}+(\frac{7}{10})^2$

$x^2+\frac{7x}{5}+\frac{49}{100}=\frac{120+49}{100}$

$x^2+\frac{7x}{5}+(\frac{7}{10})^2=\frac{169}{100}$

The left-hand side is a perfect square, hence,
$(x+\frac{7}{10})^2=\frac{169}{100}$

Taking the square root of both sides

$\sqrt{(x+\frac{7}{10})^2}=\sqrt{\frac{169}{100}}$

$x+\frac{7}{10}=\pm\frac{13}{10}$

$x=\pm\frac{13}{10}-\frac{7}{10}$

$x=\frac{+13}{10}-\frac{7}{10}$ or $\frac{-13}{10}-\frac{7}{10}$

$x=\frac{13-7}{10}$ or $\frac{-13-7}{10}$

$x=\frac{6}{10}$ or $\frac{-20}{10}$

$x=0.6$ or $-2$

Example 4

Solve $2x^2+5x+3=0$ using completing the square method.

Solution

Like the previous question, the coefficient of $x^2$ is not a perfect square, so we divide through by 2.

$\frac{\require{cancel}\bcancel{2}x^2}{\require{cancel}\bcancel{2}}+\frac{5x}{2}+\frac{3}{2}=\frac{0}{2}$

$x^2+\frac{5x}{2}+\frac{3}{2}=0$

Now, let's re-arrange the equation

$x^2+\frac{5x}{2}=-\frac{3}{2}$

Adding the square half of the coefficient of x to both sides 

$x^2+\frac{5x}{2}+(\frac{5}{4})^2=-\frac{3}{2}+(\frac{5}{4})^2$

Solving the right-hand side

$x^2+\frac{5x}{2}+(\frac{5}{4})^2=-\frac{3}{2}+\frac{25}{16}$

$x^2+\frac{5x}{2}+\frac{25}{16}=\frac{-24+25}{16}$

$x^2+\frac{5x}{2}+\frac{25}{16}=\frac{1}{16}$

The left-hand side is a perfect square
$(x+\frac{5}{4})^2=\frac{1}{16}$

Taking the square root of both sides

$\sqrt{(x+\frac{5}{4})^2}=\sqrt{\frac{1}{16}}$

$x+\frac{5}{4}=\pm\frac{1}{4}$

$x=\pm\frac{1}{4}-\frac{5}{4}$

$x=\pm0.25-1.25$

$x=0.25-1.25$ or $-0.25-1.25$

$x=-1$ or  $-1.5$

Completing the square in quadratic equations is extremely important. 

Completing the square is used to derive many formulas, including the quadratic formula.

In my next post, you will see some examples that will solidify your knowledge of quadratic equations, do well to check it out here.

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