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### Completing the Square

- You can solve quadratic equations by taking the square root of both sides of the equation. To do this, the quadratic expression must be a perfect square.
- Complete the square to make the quadratic expression into a perfect square. Take one half of the coefficient of the linear term, square it, and add it to both sides of the equation.
- If the coefficient of the quadratic term is not 1, you must first factor this coefficient out of the quadratic and linear term and then complete the square for the new quadratic expression which now has a coefficient of 1.

### Completing the Square

^{2}− 10x + 25 = 11

- Notice how the left side is a perfect square trinomial, therefore, no factoring is needed.
- x
^{2}− 10x + 25 = 11 = (x − 5)^{2}= 11 - Take the square root of both sides
- √{(x − 5)
^{2}} = ±√{11} - Simplify
- (x − 5) = ±√{11}

^{2}+ 4x + 4 = 3

- Notice how the left side is a perfect square trinomial, therefore, no factoring is needed.
- x
^{2}+ 4x + 4 = 3 = (x + 2)^{2}= 3 - Take the square root of both sides
- √{(x + 2)
^{2}} = ±√3 - Simplify
- (x + 2) = ±√3

^{2}+ 12x

- You are looking for a number c such that x
^{2}+ 12x + c is a perfect square trinomial. - Recall that whenever you have a perfect square trinomial in standard form x
^{2}+ bx + c,c = [(b^{2})/4]. - You can then use this information to factor completely x
^{2}+ bx + ( [b/2] )^{2}= x^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - Find c given b = 12
- c = [(b
^{2})/4] = [(12^{2})/4] = [144/4] = 36 - Factor Completely using the special rule
- x
^{2}+ 12x + 36 = ( x + [b/2] )^{2}= ( x + [12/2] )^{2}= ( x + 6 )^{2}

^{2}

^{2}+ 8x

- You are looking for a number c such that x
^{2}+ 12x + c is a perfect square trinomial. - Recall that whenever you have a perfect square trinomial in standard form x
^{2}+ bx + c,c = [(b^{2})/4]. - You can then use this information to factor completely x
^{2}+ bx + ( [b/2] )^{2}= x^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - Find c given b = 8
- c = [(b
^{2})/4] = [(8^{2})/4] = [64/4] = 16 - Factor completely using the special rule
- x
^{2}+ 8x + 16 = ( x + [b/2] )^{2}= ( x + [8/2] )^{2}= ( x + 4 )^{2}

^{2}

^{2}+ 3x

- You are looking for a number c such that x
^{2}+ 12x + c is a perfect square trinomial. - Recall that whenever you have a perfect square trinomial in standard form x
^{2}+ bx + c,c = [(b^{2})/4]. - You can then use this information to factor completely x
^{2}+ bx + ( [b/2] )^{2}= x^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - Find c given b = 3
- c = [(b
^{2})/4] = [(3^{2})/4] = [9/4] = [9/4] - Factor completely using the special rule
- x
^{2}+ 3x + [9/4] = ( x + [b/2] )^{2}= ( x + [3/2] )^{2}

^{2}

^{2}− 2x + 28 = 0

- Isolate the variables on the left side of the equation. Subtract 28 from both sides.
- x
^{2}− 2x = − 28 - Complete the square by adding [(b
^{2})/4] to both sides of the equation. - x
^{2}− 2x + [(b^{2})/4] = − 28 + [(b^{2})/4] - x
^{2}− 2x + [(( − 2)^{2})/4] = − 28 + [(( − 2)^{2})/4] - x
^{2}− 2x + [4/4] = − 28 + [4/4] - x
^{2}− 2x + 1 = − 27 - Factor the perfect square trinomial(left side) using the formula x
^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - ( x + [b/2] )
^{2}= ( x + [( − 2)/2] )^{2}= ( x − 1 )^{2} - Solve using the Square Root Property
- ( x − 1 )
^{2}= − 27 - √{( x − 1 )
^{2}} = ±√{ − 27} - (x − 1) = ±i√{27}
- Simplify and reduce square roots
- x = 1 ±i√{27} = 1 ±i√{9*3} = 1 ±i√9 √3 = 1 ±3i√3

^{2}− 18x + 82 = 0

- Isolate the variables on the left side of the equation. Subtract 82 from both sides.
- x
^{2}− 18x = − 82 - Complete the square by adding [(b
^{2})/4] to both sides of the equation. - x
^{2}− 18x + [(b^{2})/4] = − 82 + [(b^{2})/4] - x
^{2}− 18x + [(( − 18)^{2})/4] = − 82 + [(( − 18)^{2})/4] - x
^{2}− 18x + [324/4] = − 82 + [324/4] - x
^{2}− 18x + 81 = − 82 + 81 - x
^{2}− 18x + 81 = − 1 - Factor the perfect square trinomial(left side) using the formula x
^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - ( x + [b/2] )
^{2}= ( x + [( − 18)/2] )^{2}= ( x − 9 )^{2} - Solve using the Square Root Property
- ( x − 9 )
^{2}= − 1 - √{( x − 9 )
^{2}} = ±√{ − 1} - (x − 9) = ±i
- Simplify
- x = 9 ±i

^{2}+ 12x + 1 = − 10

- Isolate the variables on the left side of the equation. Subtract 1 from both sides.
- x
^{2}+ 12x = − 11 - Complete the square by adding [(b
^{2})/4] to both sides of the equation. - x
^{2}+ 12x + [(b^{2})/4] = − 11 + [(b^{2})/4] - x
^{2}+ 12x + [((12)^{2})/4] = − 11 + [((12)^{2})/4] - x
^{2}+ 12x + [144/4] = − 11 + [144/4] - x
^{2}+ 12x + 36 = − 11 + 36 - x
^{2}+ 12x + 81 = 25 - Factor the perfect square trinomial(left side) using the formula x
^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - ( x + [b/2] )
^{2}= ( x + [12/2] )^{2}= ( x + 6 )^{2} - Solve using the Square Root Property
- ( x + 6 )
^{2}= 25 - √{( x + 6 )
^{2}} = ±√{25} - (x + 6) = ±5
- Simplify
- x = − 6 ±5

^{2}+ 14x + 38 = − 10

- Isolate the variables on the left side of the equation. Subtract 38 from both sides.
- x
^{2}+ 14x = − 48 - Complete the square by adding [(b
^{2})/4] to both sides of the equation. - x
^{2}+ 14x + [(b^{2})/4] = − 48 + [(b^{2})/4] - x
^{2}+ 14x + [((14)^{2})/4] = − 48 + [((14)^{2})/4] - x
^{2}+ 14x + [196/4] = − 48 + [196/4] - x
^{2}+ 14x + 49 = − 48 + 49 - x
^{2}+ 14x + 49 = 1 - Factor the perfect square trinomial(left side) using the formula x
^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - ( x + [b/2] )
^{2}= ( x + [14/2] )^{2}= ( x + 7 )^{2} - Solve using the Square Root Property
- ( x + 7 )
^{2}= 1 - √{( x + 7 )
^{2}} = ±√1 - (x + 7) = ±1
- Simplify
- x = − 7 ±1

^{2}+ 4x − 8 = − 2

- Isolate the variables on the left side of the equation. Add 8 to both sides.
- 2x
^{2}+ 4x = 6 - In order to complete the square, the coefficient of x must equal 1.
- Divide everything by 2
- [(2x
^{2})/2] + [4x/2] = [6/2] = x^{2}+ 2x = 3 - Complete the square by adding [(b
^{2})/4] to both sides of the equation. - x
^{2}+ 2x + [(b^{2})/4] = 3 + [(b^{2})/4] - x
^{2}+ 2x + [((2)^{2})/4] = 3 + [((2)^{2})/4] - x
^{2}+ 2x + [4/4] = 3 + [4/4] - x
^{2}+ 2x + 1 = 3 + 1 - x
^{2}+ 2x + 1 = 4 - Factor the perfect square trinomial(left side) using the formula x
^{2}+ bx + [(b^{2})/4] = ( x + [b/2] )^{2} - ( x + [b/2] )
^{2}= ( x + [2/2] )^{2}= ( x + 1 )^{2} - Solve using the Square Root Property
- ( x + 1 )
^{2}= 4 - √{( x + 1 )
^{2}} = ±√4 - (x + 1) = ±2
- Simplify
- x = − 1 ±2

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Completing the Square

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Square Root Property 0:12
- Example: Perfect Square
- Example: Perfect Square Trinomial
- Completing the Square 4:39
- Constant Term
- Example: Complete the Square
- Solve Equations 6:42
- Add to Both Sides
- Example: Complete the Square
- Equations Where a Not Equal to 1 10:58
- Divide by Coefficient
- Example: Complete the Square
- Complex Solutions 14:05
- Real and Imaginary
- Example: Complex Solution
- Example 1: Square Root Property 18:31
- Example 2: Complete the Square 19:15
- Example 3: Complete the Square 20:40
- Example 4: Complete the Square 23:56

### Algebra 2

### Transcription: Completing the Square

*Welcome to Educator.com.*0000

*In previous lessons, we talked about solving quadratic equations through factoring and through graphing.*0002

*Today, we are going to learn another method called completing the square.*0008

*And you may have learned this earlier on in Algebra I, so it may be review.*0012

*And if you need even more depth, you can look at the Algebra I series here at Educator.com.*0017

*Now, recall that some quadratic equations can be solved by taking the square root of both sides of the equation.*0023

*And this includes equations where the quadratic expression is a perfect square.*0031

*So, for example, if you have x ^{2} - 8x + 16 = 5, you may recognize that this is a perfect square.*0037

*This trinomial is a perfect square trinomial.*0050

*It is actually equal to (x - 4) ^{2}, or (x - 4) (x - 4).*0056

*So, if I put this in this form and write it as (x - 4) ^{2} = 5, I can then solve it by taking the square root of both sides.*0063

*So, take the square root of both sides of the equation to get x - 4 = ±√-5.*0072

*And recall that you need to take both the positive and negative square root of 5,*0094

*because, thinking back, if you had something such as x ^{2} = 4,*0101

*and you were to take the square root of both sides, that would give you that x could equal 2,*0108

*or x could equal -2, because 2 ^{2} equals 4, and (-2)^{2} equals 4.*0114

*So, it is the same concept here; it looks more complex, but it is the same concept.*0122

*On the left, I am taking the square root, and then on the right, I am taking the square root of a number;*0126

*and it could be both positive square root of 5, or the negative whatever-the-square-root-of-5-turns-out-to-be.*0131

*And I can't simplify that any further without a calculator; I can just leave it in this form, ±√5.*0138

*Since this is an irrational number, I can't get an exact value; I just need to leave it in this format.*0146

*So, this is going to give me x = 4 ± √5.*0155

*And I could put this in this form; I could leave it; or I could say x = 4 + √5, and x = 4 - √5.*0159

*These are both solutions to this quadratic equation.*0172

*Now, how do you recognize perfect squares?*0176

*If you have a perfect square trinomial, and it is in the form ax ^{2} + bx + c,*0181

*then what you can do is realize that the constant equals b divided by 2, squared; or sometimes we just say it is b ^{2} over 4.*0196

*So, if you are looking, and you are not sure if it is a perfect square trinomial, look at the constant term.*0211

*And here, I have a = 1 for a coefficient; b = -8; and c = 16.*0216

*So, what I want to do is see if b ^{2}/4 equals 16.*0224

*And if I look, I see this (-8) ^{2}/4, and that equals 64 divided by 4, which is 16.*0231

*So, this checks out; so if you are not sure if you have a perfect square trinomial, you can just look at the constant.*0247

*If the rest of it checks out as perfect squares, and then you are not sure,*0252

*then you can check and say, "OK, does the constant equal b ^{2}/4? Yes, it does."*0258

*So, I knew I had a perfect square trinomial; and that makes it easy to find the square root of both sides and find the two solutions for this quadratic equation.*0266

*OK, sometimes you don't have a perfect square trinomial; in that case, what you can do is actually complete the square.*0277

*You can make it into a perfect square.*0286

*Now, what I just said is that the constant term in a perfect square trinomial equals b squared over 2, or I like to just use it as b ^{2}/4.*0288

*But it is the same thing either way.*0299

*So, for example, let's say I am given x ^{2} - 8x, similar to the last one (but that one was already a perfect square).*0303

*So, I am just given this portion; and I want to complete the square.*0313

*So, in order to complete the square, what I need is a constant term.*0320

*And I know that, in a perfect square trinomial, the constant is going to equal b ^{2}/4.*0327

*So, really what I am looking for is this.*0333

*Here, a equals 1; b equals -8; therefore, I have x ^{2} - 8x + -8^{2}/4.*0339

*x ^{2} - 8x + 8 times 8...that is 64...over 4 gives me x^{2} - 8x + 16.*0356

*And now, I have a perfect square trinomial.*0368

*So, last time I talked about if you had something and you thought it was a perfect square trinomial;*0373

*you could check it out by seeing if the constant is equal to this.*0379

*Conversely, if you don't have a perfect square, and you need to complete the square--*0385

*you just have this part, and you want to get the constant term to complete the square--*0388

*you can use this knowledge to find what that constant should be; and then, from there, we can solve equations.*0393

*Now, to solve any quadratic equation, you can complete the square by adding this, which is equal to b ^{2}/4, to both sides of the equation.*0402

*In the last slide, I talked about completing the square; but if you have an equation, you need to keep the equation balanced.*0416

*So, you can't just complete the square, and then do nothing to the other side.*0421

*So, let's take an example here, slightly different than the other one, but related.*0425

*This time, I have x ^{2} + 6x + 8 = 0; so now I have an equation.*0431

*Now, the first thing to do is: if you are dealing with an equation, get the constants on one side.*0440

*I am going to get the constants on the right, because I want to clear out*0450

*and just have these x variable terms here on the left, so I can complete the square on that.*0454

*I am going to subtract 8 from both sides to get x ^{2} + 6x = -8.*0463

*Now, I want to complete the square, and I know that, to complete the square, I am going to need x ^{2} + 6x;*0476

*but I need a constant term, and the constant term is going to be b ^{2}/4.*0483

*Now, I have to add to both sides; add b ^{2}/4 to both sides.*0491

*This is the step that sometimes gets forgotten, and then you get an incorrect answer.*0501

*OK, this is going to give me x ^{2} + 6x +...well, b is 6; now, I have to add that over here, as well; OK.*0506

*This is going to give me 36 divided by 4, which comes out to 9.*0533

*So, coming up here, x ^{2} + 6x + 9 = -8 + 9; or x^{2} + 6x + 9 = 1.*0541

*Now, I have a perfect square trinomial, and I can go and use techniques used previously in order to solve this--*0560

*such as taking the square root of both sides.*0569

*(x + 3) ^{2} (which is the same as this trinomial) equals 1.*0575

*Now, we take the square root of both sides to solve this to get x + 3 = ±√1.*0581

*So, the square root of 1 is 1, so I get x + 3 = ±1.*0590

*From there, you can just solve; I get x + 3 = 1, so x = -3 + 1; x = -2;*0598

*and I also have x + 3 = -1; therefore, I am going to get x = -4.*0609

*OK, so I am really focusing just on these first steps; but to finish it out, what you ended up with is x = -2 and x = -4 as the solutions.*0619

*To solve a quadratic equation when you don't already have a perfect square trinomial involved,*0629

*get the constants on the right and the variables on the left.*0634

*Then, complete the square by adding b ^{2}/4 to both sides.*0637

*Once you have done that, you have a perfect square here on the left.*0642

*Turn this into this form, and then take the square root of both sides to get your solutions.*0646

*Sometimes, the coefficient of x ^{2}--the leading coefficient--is not 1.*0658

*So far, we have been working with situations where we just had x ^{2}.*0664

*If the coefficient is not 1, you need to take an extra step,*0668

*and you need to divide both sides of the equation by the coefficient*0671

*in front of the x ^{2} term in order to make that coefficient 1.*0675

*Then, you just go on and complete the square as usual.*0680

*For example, if I was given 2x ^{2} - 12x + 4 = 0, I need to make this coefficient 1.*0684

*So, I can do that by dividing both sides by 2.*0694

*And that is going to cancel out right here to give me x ^{2} - 6x + 2 = 0.*0704

*OK, now I am going to go ahead and complete the square.*0714

*Remember that my first step is to get the x variables on the left, and I am going to subtract 2 from both sides to get the constants on the right.*0719

*Then, I need to complete the square: x ^{2} - 6x...I need to now add b^{2}/4 to both sides to complete the square.*0730

*This is x ^{2} - 6x + b^{2}/4 equals -2 + b^{2}/4.*0743

*OK, x ^{2} - 6x...well, b is -6, so that is + (-6)^{2}/4...= -2 + (-6)^{2}/4.*0755

*This gives me x ^{2} - 6x; this is 36 divided by 4; and I am just going to simplify that to 9.*0770

*And if I had not added to both sides, the equation would not be balanced; I would not get the correct solutions.*0781

*OK, I am coming up here to get x ^{2} - 6x + 9 = 7.*0786

*Well, now I have a perfect square trinomial, which is actually (x - 3) (this is a negative right here) squared equals 7.*0794

*Remember to take the square root of both sides to get x - 3 = ±√7.*0802

*Isolate the x to get x = 3 ± √7.*0810

*So, the two solutions are x = 3 + √7 and x = 3 - √7.*0815

*OK, so it is the same as what we just did with completing the square, and then taking the square root of both sides;*0823

*but with an extra step, because in order to complete the square, you need for the coefficient x to be 1.*0828

*If it is not 1, if you have a leading coefficient that is not 1, begin by dividing both sides by that coefficient.*0833

*OK, sometimes the solutions to a quadratic equation may be complex numbers.*0844

*Just to review: complex numbers consist of a real part and an imaginary part, something such as 3 + 2i.*0851

*So, this is in the form a + bi, where a is the real part, and 2i is the imaginary part.*0857

*It turns out that, for quadratic equations, solutions may end up in this form.*0867

*Let's look at a situation where that could occur.*0872

*Let's say I am given x ^{2} + 2x + 10 = 0.*0875

*And I am going to solve this by completing the square.*0882

*So first, I am going to isolate these x variables on the left and the constants on the right by subtracting 10 from both sides.*0885

*Now, I need to complete the square.*0892

*I want to add b ^{2}/4 to both sides of this equation.*0899

*Well, b is 2, so that is going to give me 2 ^{2}/4 = -10 + 2^{2}/4.*0910

*So, this is x ^{2} + 2x...well, 2 times 2 is 4; divided by 4--that is just 1.*0925

*So, this is -10 + 1; this gives me x ^{2} + 2x + 1 = -9.*0932

*So, I now have a perfect square trinomial; and because I have that, I will just go ahead my usual way and take the square root of both sides.*0942

*So, let's come up here to finish this and take the square root of both sides, and we rewrite it.*0951

*This is a perfect square: it is (x + 1) ^{2} = -9.*0964

*The square root of both sides gives me x + 1 = √-9.*0970

*Now, before learning about complex numbers, we would have had to stop here and say,*0975

*"OK, this is undefined; there is no real number solution; we don't know what to do with this."*0978

*But now that we have discussed imaginary numbers, we do know what to do with this.*0984

*So, let's look at how we can handle this.*0987

*We know that this equals √-1 times 9; and remember the positive and negative results for this: you need to take both: ±√-9.*0994

*OK, using the product property, this is going to equal this.*1009

*Recall that, with complex numbers, the square root of -1 is equal to i.*1018

*OK, so this is going to give me x + 1 = ± i times √9, which is 3.*1036

*Therefore, now this is something I can work with: x = (I am going to subtract -1) ± 3i.*1051

*All right, so here we have a situation where the solution to the quadratic equation is a complex number.*1061

*And the reason is: you are taking the square roots.*1067

*When you are taking the square roots, if you end up with a negative number and you take the square root of that, you are going to get an imaginary number.*1070

*So here, I ended up getting all the way down by my usual method, completing the square,*1078

*then taking the square root of both sides; and I ended up with this.*1083

*But I recognized that the square root of -9 is equal to 3i; so then, I could simply say that my solutions are x = -1 + 3i, and x = -1 - 3i.*1086

*And that is a set of complex conjugates.*1107

*So, let's use the square root property to solve this.*1112

*I am recognizing that this is a perfect square trinomial; so I don't have to complete the square--it is already done for me.*1115

*Since this is negative here, this is (x - 3) ^{2} = 7.*1123

*And if I take the square root of both sides, I am going to get x - 3 = ±√7, so x = 3 ± √7.*1127

*Or, you could write this out as x = 3 + √7, and x = 3 - √7, as the solutions.*1141

*When it is already a perfect square trinomial, it is much easier.*1151

*In this example, I am going to have to complete the square.*1156

*And this is just asking me to complete the square of x ^{2} - 10x.*1161

*So, in order to do that, I need a constant.*1166

*And for a perfect square trinomial, the constant is going to equal b ^{2}/4, so that is x^{2} - 10x + b^{2}/4.*1169

*And since this is in standard form, that is ax ^{2} + bx + c.*1183

*That means that b equals -10; so x ^{2} - 10x + (-10)^{2}, all divided by 4.*1189

*OK, so it is x ^{2} - 10x +...-10 squared is 100, divided by 4; x^{2} - 10x + 25.*1203

*This is a perfect square; this is the same as (x - 5) ^{2}.*1217

*So, to complete the square, you need to find a constant term; and the constant term is going to be equal to b ^{2}/4.*1223

*So, b equals -10; substituting in here gave me a perfect square trinomial.*1232

*Solve by completing the square: the first step is to isolate the x variables on the left side of the equation.*1241

*So, subtract 12 from both sides.*1251

*Then, complete the square by adding b ^{2}/4 to both sides; and it is very important that you do this to both sides, to keep the equation balanced.*1257

*OK, b is 6, so this gives me x ^{2} + 6x + 6^{2}/4 = -12 + 6^{2}/4, which is 36/4; 36/4 is just 9.*1271

*We will find a bit more on the right; now I have my perfect square trinomial on the left.*1306

*So, I am going to come up here and just rewrite that, because I was asked to do more than just complete the square; I actually have to solve the equation.*1321

*Now, I am going to write this as (x + 3) ^{2} = -3.*1331

*In order to solve this, I am going to take the square root of both sides.*1337

*x + 3 = ±√-3; from previous lessons, I know that I can change this to -1 times the square root of 3.*1342

*And recall that i is equal to the square root of -1, so this is i√3.*1359

*OK, this gives me x = -3 ± i√3; or I can write it out as x = -3 + i√3, and x =....*1369

*These are both solutions; they are not real-number solutions, but they are solutions.*1385

*Solving by completing the square means isolating the x variable terms on the left and the constants on the right.*1393

*Adding b ^{2}/4 to both sides gave me a perfect square trinomial on the left, which was (x + 3)^{2}, and -3 on the right.*1404

*Solving by taking the square root of both sides gave me this.*1414

*I then simplified this, since the square root of -1 is i, into i√3; and then, I isolated x to get x = -3 + i√3, x = -3 - i√3.*1419

*Again, we are going to solve by completing the square.*1437

*And my first step is always to get the variable terms on the left and the constants on the right, then complete the square.*1441

*Subtract 18 from both sides; now, here, before I go any farther, I have to look and see that my x ^{2} term is not 1.*1451

*So, if the coefficient of x ^{2} is not 1, divide both sides of the equation...*1471

*And the reason is that I need to do that in order to complete the square and go on as usual.*1488

*So, I am going to divide both sides here (and I could have done it up here, but it is easier to do it at this point, when I am simplified a bit) by 3.*1495

*These cancel out, and now my x ^{2} coefficient is 1.*1507

*12 divided by 3 gives me -4x; and 6 divided by 3 is 2.*1511

*Now, I can go about completing the square, because my leading coefficient is 1.*1517

*I am doing that by adding b ^{2}/4 to both sides: b is -4, so this gives me x^{2} - 4x, and this is...*1532

*4 times 4 is 16, divided by 4; so that is just 4; and this is also going to be 4.*1555

*OK, coming up here to finish, this is going to give me x ^{2} - 4x + 4 = 6.*1563

*I now have my perfect square trinomial here, and it is (x - 2) ^{2}; it is a negative 2, because the middle term is negative.*1576

*And I am going to find the square root of both sides; so x equals 2 plus or minus the square root of 6.*1585

*This problem took an extra step: after isolating my variables on the left and my constants on the right,*1597

*I saw that my leading coefficient was not 1, so I had to divide both sides of the equation by that coefficient 3.*1604

*Once I was there, then I went about completing the square by adding this term to both sides,*1611

*getting a perfect square trinomial (which is this), and taking the square root of both sides to get my solutions.*1618

*Thanks for visiting Educator.com; and I will see you for the next lesson.*1626

1 answer

Last reply by: Dr Carleen Eaton

Wed Nov 6, 2013 12:48 AM

Post by Chateau Siqueira on September 27, 2013

Thank you for your lectures Dr. Eaton. My college algebra class is vary fast paced which sometimes I do not absorb all the material but then I come over here an it all makes sense! I appreciate your time.

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Post by julius mogyorossy on September 1, 2013

It seems to me that one of your solutions may not be a negative #, that is what I thought at first, that your solutions may not even be the same number, one positive, one negative, a simple example, (x-2)'2=4, it is easy hear to see what the solutions are, 4 and 0. If you worked out this problem it would be, x=2+-^4, the square root of 4 is 2 so that would be 2+2=4, 2-2=0, x=4, and x=0. But don't take my word on it, ask Dr. Carleen. (x-2)'2=4 is a question, asking you what two values for x in that equation =4.

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Post by julius mogyorossy on September 1, 2013

I meant to say (x-4)'2=5 reminds me of absolute value equations. I really understand what this is saying now, x=4+-^5 is just another way to say the same thing.

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Post by julius mogyorossy on September 1, 2013

The (x-4)'2=5 reminds me of inequalities.

1 answer

Last reply by: Dr Carleen Eaton

Tue Jun 4, 2013 8:10 PM

Post by Kavita Agrawal on June 3, 2013

At about 6 min., you said that -8^2 = 64. -8^2, however, equals -64 (because of the order of operations, and exponents come before multiplying.) I think that part would make more sense if it had parentheses around it.

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Post by julius mogyorossy on March 25, 2012

It seems that if you can divide, b, in to two equal parts, getting whole numbers, you then just multiply those two equal parts times each other to get the constant in the perfect square trinomial, the first operator can be positive or negative, the second one in the perfect square trinomial will always be positive, is this correct.

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Post by Ken Mullin on January 27, 2012

Nice review of completing thr square...

Espcially like the emphasis on ISOLATING (b^2)/4 and adding result to both sides.

Some textbooks use the reciprocal of 2 and so makes the operation seem more difficult than it otherwise is...