10G (2.8) Solving quadratics by completing the square

Presentation

•

Mathematics

•

10th Grade

•

Practice Problem

•

Medium

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CCSS

HSA-REI.B.4B, HSA.REI.A.2, 8.EE.A.2

Standards-aligned

Asher Katz

Used 1+ times

FREE Resource

9 Slides • 48 Questions

10G (2.8)
Solving Quadratics by Completing the Square

When we try to solve for the values of x by "Completing the Square", we're trying to factor the polynomial into a perfect square.
An example of factoring into a perfect square: x² + 2x + 1 = ( x + 1 )( x + 1 ) = ( x + 1 )² .
The thing to remember is that our polynomial IS NOT a perfect square.
We just add a particular amount to both sides which makes it a perfect square.
In other words, we are "completing the square" .

Doing this makes it easier to solve and, by the end, the thing we added will be gone, leaving only the correct answer.
I believe that this method always works, and because of that, it can give you some pretty hairy answers

Take note that either all of the black boxes are plus or all of the black boxes are minus

Multiple Choice

Trying it out is always easier.

Step 1: Here's our polynomial

$x^2+4x+2=0$

Step 2: subtract the c term from both sides

$x^2+4x=-2$

$x^2+4x-2=-2$

$x^2+4x=0$

$x^2+4x+2=-2$

Multiple Choice

Here's the polynomial now:

$x^2+4x=-2$

Step 3: We're going to add the square completing term to both sides.

Square completing term = $\left(\frac{b}{2}\right)^2$

$x^2+4x\ +\ \left(\frac{4}{2}\right)^2=-2+\left(\frac{4}{2}\right)^2$

$x^2+4x\ =-2+\left(\frac{4}{2}\right)^2$

$x^2+4x\ +\ \left(\frac{4}{2}\right)^2=-2$

Multiple Choice

Here's the polynomial now:

$x^2+4x\ +\ \left(2\right)^2=-2+\left(2\right)^2$

Step 4: We're going to factor the left side of the equation

Factor the following; $x^2+4x\ +\ 4$

$\left(x+2\right)\left(x+2\right)$

$\left(x+4\right)\left(x+4\right)$

$\left(x+3\right)\left(x+3\right)$

$\left(x+1\right)\left(x+1\right)$

Multiple Choice

Here's the polynomial now:

$\left(x+2\right)\left(x+2\right)=-2+4$

Step 5: We're going to add up the stuff on the right side of the equation

-2 + 4 = 2

-2 + 4 = 4

-2 + 4 = 6

-2 + 4 = -6

Multiple Choice

Step 6: Put the factored left side and the added up right side together

$\left(x+2\right)\left(x+2\right)=2$

$\left(x+2\right)\left(x+2\right)+2=0$

$\left(x+2\right)=2$

$\left(x+3\right)\left(x+3\right)=5$

Multiple Choice

Here's the polynomial now, written as a square:

$\left(x+2\right)^2=2$

Step 7: We're going to take the square root of both sides of the equation

$\sqrt[]{\left(x+2\right)^2}=\sqrt[]{2}$

$\sqrt[]{\left(x+2\right)}=\sqrt[]{2}$

$\sqrt[]{x}=\sqrt[]{2}$

Multiple Choice

Here's the polynomial now:

$\sqrt[]{\left(x+2\right)^2}=\sqrt[]{2}$

Step 8: Can we make anything simpler?

[HINT: square roots cancels a squared]

$\left(x+2\right)^{ }=\pm\sqrt[]{2}$

$\sqrt[]{\left(x+2\right)}=\pm\sqrt[]{2}$

$x=\pm\sqrt[]{2}$

$\left(x+2\right)^{ }=\pm1$

Multiple Choice

$\pm\sqrt[]{2}$ can't get any simpler so here's the polynomial now:

$x+2=\pm\sqrt[]{2}$

Step 9: Subtract the 2 from both sides

$x=\pm\sqrt[]{2}-2$

$x-2=\pm\sqrt[]{2}-2$

$x+2=\pm\sqrt[]{2}-2$

$x=\pm\sqrt[]{2-2}$

Multiple Select

So we got $x=\pm\sqrt[]{2}-2$ as our answer.

Select the two values of x?

$x=+\sqrt[]{2}-2$

$x=-\sqrt[]{2}-2$

$x=-2$

$x=0$

$x=-0.5857$

Multiple Choice

Trying it out is always easier.

Step 1: Here's our polynomial:

$x^2-6x-9=0$

Step 2: subtract the c term from both sides

$x^2-6x=9$

$x^2-6x=-9$

$x^2-6x+9=9$

$x^2-6x=0$

Multiple Choice

Here's the polynomial now:

$x^2-6x=9$

Step 3: We're going to add the square completing term to both sides.

Square completing term = $\left(\frac{b}{2}\right)^2$

$x^2-6x+\left(\frac{-6}{2}\right)^2=9+\left(\frac{-6}{2}\right)^2$

$x^2-6x+\left(\frac{-6}{2}\right)^2=9\$

$x^2-6x=9\ +\ \left(\frac{-6}{2}\right)^2$

Multiple Choice

Here's the polynomial now:

$x^2-6x+\left(\frac{-6}{2}\right)^2=9+\left(\frac{-6}{2}\right)^2$

Step 4: We're going to factor the left side of the equation

Factor the following; $x^2-6x+9$

$\left(x-3\right)\left(x-3\right)$

$\left(x-4\right)\left(x-4\right)$

$\left(x-2\right)\left(x-2\right)$

$\left(x-1\right)\left(x-1\right)$

Multiple Choice

Here's the polynomial now:

$\left(x-3\right)\left(x-3\right)=9+9$

Step 5: We're going to add up the stuff on the right side of the equation

9 + 9 = 18

9 + 9 = 81

9 + 9 = 11

9 + 9 = 20

Multiple Choice

Step 6: Put the factored left side and the added-up right side back together

$\left(x-3\right)\left(x-3\right)=18$

$\left(x-3\right)\left(x-3\right)=0$

$\left(x-3\right)=18$

$\left(x+3\right)\left(x+3\right)=0$

Multiple Choice

Here's the polynomial now, written as a square:

$\left(x-3\right)^2=18$

Step 7: We're going to take the square root of both sides of the equation

$\sqrt[]{\left(x-3\right)^2}=\sqrt[]{18}$

$\sqrt[]{\left(x-3\right)}=\sqrt[]{18}$

$\sqrt[]{x}=\sqrt[]{18}$

Multiple Choice

Here's the polynomial now:

$\sqrt[]{\left(x-3\right)^2}=\sqrt[]{18}$

Step 8: Can we make anything simpler?

[HINT: square roots cancels a squared]

$x-3=\pm\sqrt[]{18}$

$\sqrt[]{\left(x-3\right)}=\pm\sqrt[]{18}$

$x=\pm\sqrt[]{18}$

$\left(x-3\right)^{ }=\pm1$

Multiple Choice

$\pm\sqrt[]{18}$ can't get any simpler so here's the polynomial now:

$x-3=\pm\sqrt[]{18}$

Step 9: Add 3 to both sides

$x-3+3=\pm\sqrt[]{18}+3$

$x-3=\pm\sqrt[]{18}-3$

$x+3=\pm\sqrt[]{18}+3$

$x=\pm\sqrt[]{18+3}$

Multiple Select

So we got $x=\pm\sqrt[]{18}+3$ as our answer.

Select the two values of x?

$x=+\ \sqrt[]{18}+3$

$x=-\ \sqrt[]{18}+3$

$x=-18$

$x=0$

$x=7.242640$

Multiple Choice

Trying it out is always easier.

Step 1: Here's our polynomial:

$2x^2+8x-8=0$

Step 1.5: We can simplify the polynomial first by factoring out a 2

$2\left(x^2+4x-4\right)=0$

$2\left(x^2+8x-8\right)=0$

$2\left(x^2+4x-2\right)=0$

Multiple Choice

So we can actually divide both sides of the equation by 2 so here's our polynomial now:

$x^2+4x-4=0$

Step 2: subtract the c term from both sides

$x^2+4x=4$

$x^2+4x=-4$

$x^2+4x=0$

$x^2+4x-4=0$

Multiple Choice

Here's the polynomial now:

$x^2+4x=4$

Step 3: We're going to add the square completing term to both sides.

Square completing term = $\left(\frac{b}{2}\right)^2$

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

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

Multiple Choice

Here's the polynomial now:

$\left(\frac{4}{2}\right)^2=4\ \ \ so\ \ \ \ x^2+4x+4=4+4$

Step 4: We're going to factor the left side of the equation

Factor the following; $x^2+4x+4$

$\left(x+2\right)\left(x+2\right)$

$\left(x+4\right)\left(x+4\right)$

$\left(x+3\right)\left(x+3\right)$

$\left(x-1\right)\left(x-1\right)$

Multiple Choice

Here's the polynomial now:

$\left(x+2\right)\left(x+2\right)=4+4$

Step 5: We're going to add up the stuff on the right side of the equation

4 + 4 = 8

4 + 4 = 6

4 + 4 = 4

4 + 4 = 16

Multiple Choice

Step 6: Put the factored left side and the added-up right side back together

$\left(x+2\right)\left(x+2\right)=8$

$\left(x+2\right)\left(x+2\right)=0$

$\left(x+2\right)=8$

$\left(x+3\right)\left(x+3\right)=8$

Multiple Choice

Here's the polynomial now, written as a square:

$\left(x+2\right)^2=8$

Step 7: We're going to take the square root of both sides of the equation

$\sqrt[]{\left(x+2\right)^2}=\sqrt[]{8}$

$\sqrt[]{\left(x+2\right)}=\sqrt[]{8}$

$\sqrt[]{x}=\sqrt[]{8}$

Multiple Choice

Here's the polynomial now:

$\sqrt[]{\left(x+2\right)^2}=\sqrt[]{8}$

Step 8: Can we make anything simpler?

[HINT: square roots cancels a squared]

$x+2=\pm\sqrt[]{8}$

$\sqrt[]{\left(x+2\right)}=\pm\sqrt[]{8}$

$x=\pm\sqrt[]{8}$

$\left(x+2\right)^{ }=\pm1$

Multiple Choice

$\pm\sqrt[]{8}$ can't get any simpler so here's the polynomial now:

$x+2=\pm\sqrt[]{8}$

Step 9: Subtract 2 from both sides

$x+2-2=\pm\sqrt[]{8}-2$

$x-2=\pm\sqrt[]{8}-2$

$x+2=\pm\sqrt[]{8}-2$

$x=\pm\sqrt[]{8-2}$

Multiple Select

So we got $x=\pm\ \sqrt[]{8}-2$ as our answer.

Select the two values of x?

$x=+\ \sqrt[]{8}-2$

$x=-\ \sqrt[]{8}-2$

$x=-8$

$x=0$

$x=-0.828427$

Multiple Choice

Step 1: here's our polynomial:

$2x^2-8x+9=0$

Step 2: subtract the c term from both sides

$2x^2-8x=-9$

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

$2x^2-8x-9=-9$

$2x^2-8x=0$

Multiple Choice

Here's our polynomial now:

$2x^2-8x=-9$

Step 2.5: We can actually simplify the left side of the polynomial first.

$2\left(x^2-4x\right)=-9$

$x^2-4x+9=-9$

$2\left(x^2-4x-5\right)=-9$

$2\left(x^2-4x\right)=0$

Multiple Choice

Here's the polynomial now:

$2\left(x^2-4x\right)=-9$

Step 3: We're going to add the square completing term to both sides.

[HINT: We're trying to complete the square in the parenthesis on the left]

Square completing term = $\left(\frac{b}{2}\right)^2$

$2\left(x^2-4x\ +\left(\frac{4}{2}\right)^2\right)=-9\ +\left(\frac{4}{2}\right)^2$

$2\left(x^2-4x\right)=-9+\left(\frac{4}{2}\right)^2$

$2\left(x^2-4x+\left(\frac{4}{2}\right)^2\right)=-9$

Multiple Choice

Here's the polynomial now:

$\left(\frac{4}{2}\right)^2=4\ \ \ \ so\ \ 2\left(x^2-4x\ +4\right)=-9\ +4$

Step 4: We're going to factor inside the parenthesis on the left side of the equation

Factor the following; $x^2-4x+4$

$\left(x-2\right)\left(x-2\right)$

$\left(x-4\right)\left(x-4\right)$

$\left(x-3\right)\left(x-3\right)$

$\left(x-1\right)\left(x-1\right)$

Multiple Choice

Here's the polynomial now:

$\left(x-2\right)\left(x-2\right)=-9+4$

Step 5: We're going to add up the stuff on the right side of the equation

-9 + 4 = -5

-9 + 4 = 4

-9 + 4 = -13

-9 + 4 = 13

Multiple Choice

Step 6: Put the factored left side and the added-up right side back together

$\left(x-2\right)\left(x-2\right)=-5$

$\left(x-2\right)\left(x-2\right)=0$

$\left(x-2\right)=-5$

$\left(x-4\right)\left(x-4\right)=-5$

Multiple Choice

Here's the polynomial now, written as a square:

$\left(x-2\right)^2=-5$

Step 7: We're going to take the square root of both sides of the equation

$\sqrt[]{\left(x-2\right)^2}=\sqrt[]{-5}$

$\sqrt[]{\left(x-2\right)}=\sqrt[]{-5}$

$\sqrt[]{x}=\sqrt[]{-5}$

Multiple Choice

Here's the polynomial now:

$\sqrt[]{\left(x-2\right)^2}=\sqrt[]{-5}$

Step 8: Can we make anything simpler?

[HINT: square roots cancels a squared]

$x-2=\pm\sqrt[]{-5}$

$\sqrt[]{\left(x-2\right)}=\pm\sqrt[]{-5}$

$x=\pm\sqrt[]{-5}$

$\left(x-2\right)^{ }=\pm5$

Multiple Choice

$\pm\sqrt[]{-5}$ can't get any simpler so here's the polynomial now:

$x-2=\pm\sqrt[]{-5}$

Step 9: Add 2 to both sides

$x+2-2=\pm\sqrt[]{-5}+2$

$x-2=\pm\sqrt[]{-5}-2$

$x+2=\pm\sqrt[]{-5}-2$

$x=\pm\sqrt[]{-5+2}$

Multiple Select

So we got $x=\pm\ \sqrt[]{-5}+2$ as our answer.

Select the two values of x?

$x=+\ \sqrt[]{-5}+2$

$x=-\ \sqrt[]{-5}+2$

$x=-5$

$x=0$

$x=-0.7584793$

Multiple Choice

Find the values of x:

x² + 14x − 38 = 0

$x\ =\ -7\pm\sqrt[]{87}$

x = -5 or x = 7

x = -14 or x = 24

x = -19 or x = 2

x = -10 or x = 4

Multiple Choice

Find the values of x:

x² + 14x − 51 = 0

x = 15 or x = -9

x = 10 or x = -7

x = 20 or x = -5

x = -17 or x = 3

x = 25 or x = -2

Multiple Choice

Find the values of x:

x² - 12x + 11 = 0

x = 4 or x = 8

x = 2 or x = 10

x = -3 or x = 15

x = 1 or x = 11

x = 5 or x = 7

Multiple Choice

Find the values of x:

x² - 4x - 98 = 0

$x\ =\ 2\pm\sqrt[]{102}$

x = -10 or x = 8

x = -3 or x = 15

x = -7 or x = 11

x = -1 or x = 9

Multiple Choice

Find the values of x:

x² - 10x + 26 = 8

x = -5

x = 10 $\pm\sqrt[]{5}$

x = 3

x = 7

x = 5 ± $\sqrt[]{7}$

Multiple Choice

Find the values of x:

6x² + 12x - 48 = 0

x = 0 or x = 6

x = -10 or x = 8

x = 3 or x = 5

x = 2 or x = -4

x = -4 or x = 12

Multiple Choice

Find the values of x:

5x² + 20x - 60 = 0

x = -3 or x = 4

x = 3 or x = -8

x = 2 or x = -6

x = 10 or x = -6

x = 8 or x = -10

Multiple Choice

Find the values of x:

8x² + 16x - 42 = 0

$x\ =\ \frac{3}{2},-\frac{7}{2}$

$x\ =-\frac{1}{2}\pm\sqrt[]{13}$

$x\ =4\pm\sqrt[]{13}$

x = $\pm13$

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

Multiple Choice

Find the values of x:

2x² = -6 + 8x

x = -3, x = 6

x = 1, x = 2

x = 1, x = 3

x = -2, x = 5

x = 0, x = 4

Multiple Choice

Find the values of x:

2x² - 5x + 67 = 0

[HINT: the values will be complex]

$x\ =\ \pm\ \sqrt[]{-\frac{1047}{32}}$

$x\ =\ \pm\ \sqrt[]{-1047}+5$

$x\ =\ \pm\ \sqrt[]{-\frac{1047}{32}}+\frac{5}{2}$

$x\ =\ \pm\ \sqrt[]{-32}+\frac{5}{2}$

We're going to do some more complicated polynomials before the midterm but for now, focus on this and the fact that "completing the square" always works, but the steps can sometimes be fuzzy for the harder problems, if you get stuck on a step , you need to be creative, there is probably some extra little step that you need do to make it work.

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10G (2.8) Solving quadratics by completing the square

10G (2.8) Solving Quadratics by Completing the Square

​Take note that either all of the black boxes are plus or all of the black boxes are minus

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10G (2.8)
Solving Quadratics by Completing the Square

Take note that either all of the black boxes are plus or all of the black boxes are minus