Equation Lab #19 – Quadratic Equations: Completing the Square

During the last few days, I went over how to solve quadratic equations by factoring. Now that’s been taken care of, it’s time to move to the next lesson – completing the square. What you’ll be doing here is turning the trinomial into a perfect square trinomial.

Completing the Square:

Let’s start with the equation $x^{2}=81$. Take the square root of both sides, you should get $x=9$ and $x=-9$. Mission accomplished! Now what about the equation $x^{2}+2x+1=81$? This seems complicated since you have a trinomial on the left. But don’t get up yet. Remember when I went over perfect square trinomials yesterday? $x^{2}+2x+1$ is a perfect square trinomial. If we factored it, we get $(x+1)^{2}$, so $x^{2}+2x+1=81$ is the same as $(x+1)^{2}=81$. You can now take the square root. The result is $x+1=9$ and $x+1=-9$. What about $x^{2}+2x+10=90$? That’s what today’s lesson is all about.

So completing the square is about changing the quadratic expression on the left side of the equation into a perfect square trinomial. Although this method can work on any quadratic equation for as long as a linear term exist, it’s preferred that it’s only to be used when $a=1$ and $b$ is an even integer.

To complete the square, you need to find $(-\frac{b}{2a})^{2}$. When $a=1$, you can simply use $(\frac{b}{2})^{2}$ to find the constant that builds the perfect square trinomial. Take for instance, the linear term is $6x$ and the quadratic term is $x^{2}$. What constant builds a perfect square trinomial when the linear term is $6x$ and the quadratic term is $x^{2}$? The answer is $9$, because when $b=6$, $b/2=3$, and $3^{2}=9$. But what if the constant being applied to the two variable terms is anything but $9$? You can either move the constant to both sides before completing the square, or you can change the constant into an operation of $9$ and another number. If you did the latter, you can move it over to the other side. Remember that this requires the opposite operation.

Once the left side is a perfect square trinomial and the right side is a constant term, you can factor the trinomial down into a squared binomial and take the square root of both sides. You will be left with a binomial and another number. This is when you can solve for the variable using basic means.

Let’s start with our first equation, $x^{2}+10x+16=40$. $10/2$ is $5$, and $5^{2}=25$. So the expression on the left side should be $x^{2}+10x+25$. But the constant on the left side is $16$, not $25$. What operation involving $25$ results in $16$? Since $25$ is greater than $16$, the answer is subtraction. And what number do you need to subtract from $25$ to get the difference of $16$? The answer is $9$. So $16$ can be rewritten as $25-9$. So the equation is $x^{2}+10x+25-9=40$.

Going to the next step, instead of combining like terms since that would send us back to square one, we must move the $-9$ to the other side. Since $9$ is being applied via subtraction, we must add $9$ to both sides of the equation. $x^{2}+10x+25$ is now on its own, but $40+9=49$. So the equation is $x^{2}+10x+25=49$. As we figured that $5$ is half of $10$ and the square root of $25$, you can rewrite the equation as $(x+5)^{2}=49$.

We can now take the square root of both sides. $\sqrt{(x+5)^{2}}=x+5$, but $\sqrt{49}=\pm 7$. This means that either $x+5=-7$ or $x+5=7$. Subtracting $5$ from both sides, $-7$ gets an expansion to $-12$, while $7$ is reduced to $2$. The solution is $x=-12$ or $x=2$.

1. $x^{2}+10x+16=40$
2. $10/2=5; 5^{2}=25$
3. $16=25-9$
4. $x^{2}+10x+25-9=40$
5. $x^{2}+10x+25=49$
6. $(x+5)^{2}=49$
7. $x+5=\pm7$
8. $x=-12, x=2$

Checking our work, let’s see if either solution is correct. If $x=-12$, then $x^{2}=144$ and $10x=-120$. $144-120+16=40$, so that solution holds. If $x=2$, then $x^{2}=4$ and $10x=20$. $4+20+16=40$, so that solution holds. Therefore, both $x=-12$ and $x=2$ are correct.

Now let’s try another. Our equation is $4x^{2}-12x+23=183$. Since $a\ne 1$, it’s not ideal to use this method. But $4x^{2}$ and $12x$ are part of another perfect square trinomial. The square root of $4$ is $2$. $-\frac{-12}{2\cdot 2}=3$. When you square $3$, you get $9$. So the trinomial on the left side should be $4x^{2}-12x+9$. The only dissonance we see is $9$ and $23$. What operation involving $9$ results in $23$? Since $9$ is less than $23$, the answer is addition. And the number we add to $9$ to get $23$ is $14$. The equation is now $4x^{2}-12x+9+14=183$

The next thing we do is to subtract $14$ from both sides. The equation is now $4x^{2}-12x+9=169$. Factor the left side down, we get $(2x-3)^{2}=169$. We can now take the square root of both sides. The equation becomes $2x-3=\pm 13$. This means either $2x-3=-13$ or $2x-3=13$. If it equals $-13$, adding $3$ to both sides makes the equation $2x=-10$. Divide both sides by $2$, you get $x=-5$. If it equals $13$, adding $3$ to both sides makes the equation $2x=16$. Divide both sides by $2$, you get $x=8$. Our solutions are $x=-5$ and $x=8$.

1. $4x^{2}-12x+23=183$
2. $\sqrt{4}=2; 12/(2\cdot 2)=3; 3^{2}=9$
3. $23=9+14$
4. $4x^{2}-12x+9+14=183$
5. $4x^{2}-12x+9=169$
6. $(2x-3)^{2}=169$
7. $2x-3=\pm 13$
8. $2x=-10, 2x=16$
9. $x=-5, x=8$

Checking our work, let’s see if either solution is correct. If $x=-5$, then $4x^{2}=100$ and $-12x=60$. $100+60+23=183$, so this solution holds. If $x=8$, then $4x^{2}=256$ and $-12x=-96$. $256-96+23=183$, so this solution holds as well. Therefore, both $x=-5$ and $x=8$ are correct.

Word Problem #17:

The area of a bathroom is 32 square feet. Compared to the square bedroom it’s part of, the length is 4 feet shorter while the width is 8 feet shorter. How big is the bedroom?

So let’s set up the equation. Let $l$ be the length of each wall in the bedroom. It looks like the bathroom has an area of $(l-4)(l-8)$. But we know that the bathroom has an area of 32 square feet. So the equation is $(l-4)(l-8)=32$.

First, let’s multiply the two binomials. The result should be $l^{2}-12l+32$, so our equation is $l^{2}-12l+32=32$. Since the leading coefficient is $1$, all we need to do is to divide the linear term’s coefficient, $-12$, by $2$ and square the quotient. You should get $36$. Since $36$ is greater than $32$ by $4$, we write the left side as $l^{2}-12l+36-4$. Add $4$ to both sides, you get $l^{2}-12l+36=36$.

Now we factor down the trinomial on the left side. It should become $(l-6)^{2}$, so the equation is $(l-6)^{2}=36$. Take the square root of both sides, we get $l-6=\pm 6$. Add $6$ to both sides, you get $l=0$ and $l=12$. It’s not possible for a room to have any walls be zero feet wide, so our solution is $l=12$.

 1. Define $l$ as length.
 2. Bathroom has a length that is 4 feet
    shorter and 8 feet shorter than the
    bedroom's standard lengths. The area
    is 32 square feet.
 3. $(l-4)(l-8)=32$
 4. $l^{2}-12l+32=32$
 5. $12/2=6; 6^{2}=36$
 6. $32=36-4$
 7. $l^{2}-12l+36-4=32$
 8. $l^{2}-12l+36=36$
 9. $(l-6)^{2}=36$
10. $l-6=\pm 6$
11. $l=0, l=12$

The length of a bedroom wall is 12 feet. But we’re not done yet. The real answer asks for area, not length. We still need to find the area. $12\cdot 12=144$, so the bedroom has an area of 144 square feet.

Quiz #19:

Now that the lesson is over, let’s see if you can complete this quiz. Like always, the correct solutions and the correct variables must be used. But for this quiz, I will take any answer for as long as it fits the solution. So if the answer is $a=3,-3$, you can use “$3$“, “$-3$“, “$3,-3“$, “$-\mathrm{3,3}$“, “$a=3$“, “$a=-3$“, “$a=3,-3$“, or “$a=-\mathrm{3,3}$“.