The area of another parallelogram is given by the equation 3x^2 + 9x - 5 = 0 , where x is the length of the base. What is the length of this base?
Guidance
When there is a number in front of @$x^2@$ , it will make completing the square a little more complicated. See how the steps change in Example A.
Example A
Determine the number c that completes the square of @$2x^2 - 8x + c @$ .
Solution: In the previous concept, we just added @$\left(\frac{b}{2}\right)^2@$ , but that was when @$a=1@$ . Now that @$a \neq 1@$ , we have to take the value of a into consideration. Let's pull out the GCF of 2 and 8 first.
@$2 \left(x^2 - 4x \right)@$
Now, there is no number in front of @$x^2@$ .
@$\left(\frac{b}{2}\right)^2 = \left(\frac{4}{2}\right)^2 = 4@$ .
Add this number inside the parenthesis and distribute the 2.
@$2 \left(x^2 - 4x +4 \right)=2x^2-4x+8@$
So, @$c=8@$ .
Example B
Solve @$3x^2-9x+11=0@$
Solution:
1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.
@$$3x^2-9x=-11@$$
2. Pull out @$a@$ from everything on the left side. Even if @$b@$ is not divisible by @$a@$ , the coefficient of @$x^2@$ needs to be 1 in order to complete the square.
@$$3(x^2-3x+\underline{\;\;\;\;\;\;})=-11@$$
3. Now, complete the square. Determine what number would make a perfect square trinomial.
To do this, divide the @$x-@$ term by 2 and square that number, or @$\left(\frac{b}{2}\right)^2@$ .
@$$\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}@$$
4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.
@$$3\left(x^2-3x {\color{red}+\frac{9}{4}}\right)=-11 {\color{red}+\frac{27}{4}}@$$
5. Factor the left side and simplify the right.
@$$3\left(x-\frac{3}{2}\right)^2=-\frac{17}{4}@$$
6. Solve by using square roots.
@$$\left(x-\frac{3}{2}\right)^2 &=-\frac{17}{12}\\ x-\frac{3}{2} &=\pm \frac{i\sqrt{17}}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\ x &=\frac{3}{2}\pm \frac{\sqrt{51}}{6}i@$$
Be careful with the addition of step 2 and the changes made to step 4. A very common mistake is to add @$\left(\frac{b}{2}\right)^2@$ to both sides, without multiplying by @$a@$ for the right side.
Example C
Solve @$4x^2+7x-18=0@$ .
Solution: Let’s follow the steps from Example B.
1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.
@$$4x^2-7x=18@$$
2. Pull out @$a@$ from everything on the left side.
@$$4\left(x^2+\frac{7}{4}x+\underline{\;\;\;\;\;\;}\right)=18@$$
3. Now, complete the square. Find @$\left(\frac{b}{2}\right)^2@$ .
@$$\left(\frac{b}{2}\right)^2=\left(\frac{7}{8}\right)^2=\frac{49}{64}@$$
4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.
@$$4\left(x^2+\frac{7}{4}x{\color{red}+\frac{49}{64}}\right)=18{\color{red}+\frac{49}{16}}@$$
5. Factor the left side and simplify the right.
@$$4\left(x+\frac{7}{8}\right)^2=\frac{337}{16}@$$
6. Solve by using square roots.
@$$\left(x+\frac{7}{8}\right)^2 &=\frac{337}{64}\\ x+\frac{7}{8} &=\pm \frac{\sqrt{337}}{8}\\ x &=-\frac{7}{8} \pm \frac{\sqrt{337}}{8}@$$
Intro Problem Revisit We can't factor @$3x^2 + 9x - 5 = 0@$ , so let's follow the step-by-step process we learned in this lesson.
1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.
@$$3x^2 + 9x = 5@$$
2. Pull out @$a@$ from everything on the left side.
@$$3\left(x^2 + 3x+\underline{\;\;\;\;\;\;}\right) = 5@$$
3. Now, complete the square. Find @$\left(\frac{b}{2}\right)^2@$ .
@$$\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}@$$
4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.
@$$3\left(x^2+ 3x{\color{red}+\frac{9}{4}}\right)=5{\color{red}+\frac{27}{4}}@$$
5. Factor the left side and simplify the right.
@$$3\left(x+\frac{3}{2}\right)^2=\frac{47}{4}@$$
6. Solve by using square roots.
@$$\left(x+\frac{3}{2}\right)^2 &=\frac{47}{12}\\ x+\frac{3}{2} &=\pm \frac{\sqrt{47}}{\sqrt{12}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{47}}}{2\sqrt{3}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{141}}}{6} @$$
However, because x is the length of the parallelogram's base, it must have a positive value. Only @$x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}@$ results in a positive value, so the length of the base is @$x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}@$ .
Guided Practice
Solve the following quadratic equations by completing the square.
1. @$5x^2+29x-6=0@$
2. @$8x^2-32x+4=0@$
Answers
Use the steps from the examples above to solve for @$x@$ .
1.
@$$5x^2+29x-6 &=0\\ 5\left(x^2+\frac{29}{5}x\right) &=6\\ 5\left(x^2+\frac{29}{5}x+\frac{841}{100}\right) &=6+\frac{841}{20}\\ 5\left(x+\frac{29}{10}\right)^2 &=\frac{961}{20}\\ \left(x+\frac{29}{10}\right)^2 &=\frac{961}{100}\\ x+\frac{29}{10} &=\pm \frac{31}{10}\\ x &=-\frac{29}{10} \pm \frac{31}{10}\\ x &=-6, \frac{1}{5}@$$
2.
@$$8x^2-32x+4 &=0\\ 8(x^2-4x) &=-4\\ 8(x^2-4x+4) &=-4+32\\ 8(x-2)^2 &=28\\ (x-2)^2 &=\frac{7}{2}\\ x-2 &=\pm \frac{\sqrt{7}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\ x &=2\pm \frac{\sqrt{14}}{2}@$$
Explore More
Solve the quadratic equations by completing the square.
- @$6x^2-12x-7=0@$
- @$-4x^2+24x-100=0@$
- @$5x^2-30x+55=0@$
- @$2x^2-x-6=0@$
- @$\frac{1}{2}x^2+7x+8=0@$
- @$-3x^2+4x+15=0@$
Solve the following equations by factoring, using square roots, or completing the square.
- @$4x^2-4x-8=0@$
- @$2x^2+9x+7=0@$
- @$-5(x+4)^2-19=26@$
- @$3x^2+30x-5=0@$
- @$9x^2-15x-6=0@$
- @$10x^2+40x+88=0@$
Problems 13-15 build off of each other.
- Challenge Complete the square for @$ax^2+bx+c=0@$ . Follow the steps from Examples A and B. Your final answer should be in terms of @$a, b,@$ and @$c@$ .
- For the equation @$8x^2+6x-5=0@$ , use the formula you found in #13 to solve for @$x@$ .
- Is the equation in #14 factorable? If so, factor and solve it.
- Error Analysis Examine the worked out problem below.
@$$4x^2-48x+11&=0\\ 4(x^2-12x+\underline{\;\;\;\;\;\;}) &=-11\\ 4(x^2-12x+36) &=-11+36\\ 4(x-6)^2 &=25\\ (x-6)^2 &=\frac{25}{4}\\ x-6 &=\pm \frac{5}{2}\\ x &=6\pm \frac{5}{2} \rightarrow \frac{17}{2},\frac{7}{2}@$$
Plug the answers into the original equation to see if they work. If not, find the error and correct it.