Completing The Square: Solve Quadratic Equations Easily

Hey guys! Let's dive into a super useful technique in algebra: completing the square. This method is fantastic for solving quadratic equations, and it's also a stepping stone to understanding more advanced concepts. We're going to break down three different equations step-by-step, so you'll be a pro at completing the square in no time. Let's get started!

1. Solving $x^2 - 7x + 12 = 0$ by Completing the Square

Okay, let's kick things off with the equation $x^2 - 7x + 12 = 0$. Our goal here is to rewrite this equation in a form that looks like $(x - a)^2 = b$, which makes it super easy to solve for $x$.

Step 1: Move the Constant Term

First, we want to isolate the $x^2$ and $x$ terms on one side of the equation. To do this, we'll subtract 12 from both sides:

$x^2 - 7x = -12$

Step 2: Complete the Square

Now comes the fun part: completing the square. To do this, we need to add a value to both sides of the equation that will make the left side a perfect square trinomial. The value we need to add is $(\frac{b}{2})^2$, where $b$ is the coefficient of the $x$ term. In our case, $b = -7$, so we need to add $(\frac{-7}{2})^2 = (\frac{49}{4})$ to both sides:

$x^2 - 7x + \frac{49}{4} = -12 + \frac{49}{4}$

Step 3: Factor and Simplify

The left side of the equation is now a perfect square trinomial, which we can factor as $(x - \frac{7}{2})^2$. On the right side, we need to find a common denominator to add the numbers:

$(x - \frac{7}{2})^2 = -\frac{48}{4} + \frac{49}{4}$

$(x - \frac{7}{2})^2 = \frac{1}{4}$

Step 4: Take the Square Root

Next, we take the square root of both sides of the equation. Remember to include both the positive and negative square roots:

$x - \frac{7}{2} = \pm \sqrt{\frac{1}{4}}$

$x - \frac{7}{2} = \pm \frac{1}{2}$

Step 5: Solve for x

Finally, we solve for $x$ by adding $\frac{7}{2}$ to both sides:

$x = \frac{7}{2} \pm \frac{1}{2}$

This gives us two possible solutions:

$x = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$

$x = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3$

So the solutions to the equation $x^2 - 7x + 12 = 0$ are $x = 4$ and $x = 3$.

2. Solving $x^2 - x - 5 = 0$ by Completing the Square

Alright, let's move on to our second equation: $x^2 - x - 5 = 0$. We'll follow the same steps as before, but this time we'll be dealing with slightly different numbers.

Step 1: Move the Constant Term

First, we add 5 to both sides of the equation to isolate the $x^2$ and $x$ terms:

$x^2 - x = 5$

Step 2: Complete the Square

Now we complete the square. The coefficient of the $x$ term is -1, so we need to add $(\frac{-1}{2})^2 = \frac{1}{4}$ to both sides:

$x^2 - x + \frac{1}{4} = 5 + \frac{1}{4}$

Step 3: Factor and Simplify

The left side is now a perfect square trinomial, which we can factor as $(x - \frac{1}{2})^2$. On the right side, we need to find a common denominator to add the numbers:

$(x - \frac{1}{2})^2 = \frac{20}{4} + \frac{1}{4}$

$(x - \frac{1}{2})^2 = \frac{21}{4}$

Step 4: Take the Square Root

Next, we take the square root of both sides of the equation. Remember to include both the positive and negative square roots:

$x - \frac{1}{2} = \pm \sqrt{\frac{21}{4}}$

$x - \frac{1}{2} = \pm \frac{\sqrt{21}}{2}$

Step 5: Solve for x

Finally, we solve for $x$ by adding $\frac{1}{2}$ to both sides:

$x = \frac{1}{2} \pm \frac{\sqrt{21}}{2}$

This gives us two possible solutions:

$x = \frac{1 + \sqrt{21}}{2}$

$x = \frac{1 - \sqrt{21}}{2}$

So the solutions to the equation $x^2 - x - 5 = 0$ are $x = \frac{1 + \sqrt{21}}{2}$ and $x = \frac{1 - \sqrt{21}}{2}$.

3. Solving $2x^2 - 4x - 30 = 0$ by Completing the Square

Last but not least, let's tackle the equation $2x^2 - 4x - 30 = 0$. This one has an extra twist because the coefficient of the $x^2$ term is not 1. Don't worry, we'll handle it!

Step 1: Divide by the Leading Coefficient

First, we need to divide the entire equation by the leading coefficient, which is 2:

$x^2 - 2x - 15 = 0$

Step 2: Move the Constant Term

Now, we add 15 to both sides of the equation to isolate the $x^2$ and $x$ terms:

$x^2 - 2x = 15$

Step 3: Complete the Square

Next, we complete the square. The coefficient of the $x$ term is -2, so we need to add $(\frac{-2}{2})^2 = 1$ to both sides:

$x^2 - 2x + 1 = 15 + 1$

Step 4: Factor and Simplify

The left side is now a perfect square trinomial, which we can factor as $(x - 1)^2$. On the right side, we simply add the numbers:

$(x - 1)^2 = 16$

Step 5: Take the Square Root

Next, we take the square root of both sides of the equation. Remember to include both the positive and negative square roots:

$x - 1 = \pm \sqrt{16}$

$x - 1 = \pm 4$

Step 6: Solve for x

Finally, we solve for $x$ by adding 1 to both sides:

$x = 1 \pm 4$

This gives us two possible solutions:

$x = 1 + 4 = 5$

$x = 1 - 4 = -3$

So the solutions to the equation $2x^2 - 4x - 30 = 0$ are $x = 5$ and $x = -3$.

Conclusion

And there you have it! We've successfully solved three different quadratic equations by completing the square. Remember, the key steps are to isolate the $x^2$ and $x$ terms, complete the square by adding $(\frac{b}{2})^2$ to both sides, factor the perfect square trinomial, take the square root, and solve for $x$. Keep practicing, and you'll become a completing-the-square master in no time! Happy solving! You got this! Good luck!