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Lecture Comments (9)

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.

0 answers

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.

0 answers

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.

0 answers

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.

0 answers

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.

0 answers

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...

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

Use the square root property to solve x2 − 10x + 25 = 11
  • Notice how the left side is a perfect square trinomial, therefore, no factoring is needed.
  • x2 − 10x + 25 = 11 = (x − 5)2 = 11
  • Take the square root of both sides
  • √{(x − 5)2} = ±√{11}
  • Simplify
  • (x − 5) = ±√{11}
x = 5 ±√{11}
Use the square root property to solve x2 + 4x + 4 = 3
  • Notice how the left side is a perfect square trinomial, therefore, no factoring is needed.
  • x2 + 4x + 4 = 3 = (x + 2)2 = 3
  • Take the square root of both sides
  • √{(x + 2)2} = ±√3
  • Simplify
  • (x + 2) = ±√3
x = − 2 ±√3
Complete the square x2 + 12x
  • You are looking for a number c such that x2 + 12x + c is a perfect square trinomial.
  • Recall that whenever you have a perfect square trinomial in standard form x2 + bx + c,c = [(b2)/4].
  • You can then use this information to factor completely x2 + bx + ( [b/2] )2 = x2 + bx + [(b2)/4] = ( x + [b/2] )2
  • Find c given b = 12
  • c = [(b2)/4] = [(122)/4] = [144/4] = 36
  • Factor Completely using the special rule
  • x2 + 12x + 36 = ( x + [b/2] )2 = ( x + [12/2] )2 = ( x + 6 )2
( x + 6 )2
Complete the square x2 + 8x
  • You are looking for a number c such that x2 + 12x + c is a perfect square trinomial.
  • Recall that whenever you have a perfect square trinomial in standard form x2 + bx + c,c = [(b2)/4].
  • You can then use this information to factor completely x2 + bx + ( [b/2] )2 = x2 + bx + [(b2)/4] = ( x + [b/2] )2
  • Find c given b = 8
  • c = [(b2)/4] = [(82)/4] = [64/4] = 16
  • Factor completely using the special rule
  • x2 + 8x + 16 = ( x + [b/2] )2 = ( x + [8/2] )2 = ( x + 4 )2
( x + 4 )2
Complete the square x2 + 3x
  • You are looking for a number c such that x2 + 12x + c is a perfect square trinomial.
  • Recall that whenever you have a perfect square trinomial in standard form x2 + bx + c,c = [(b2)/4].
  • You can then use this information to factor completely x2 + bx + ( [b/2] )2 = x2 + bx + [(b2)/4] = ( x + [b/2] )2
  • Find c given b = 3
  • c = [(b2)/4] = [(32)/4] = [9/4] = [9/4]
  • Factor completely using the special rule
  • x2 + 3x + [9/4] = ( x + [b/2] )2 = ( x + [3/2] )2
( x + [3/2] )2
Solve by completing the square x2 − 2x + 28 = 0
  • Isolate the variables on the left side of the equation. Subtract 28 from both sides.
  • x2 − 2x = − 28
  • Complete the square by adding [(b2)/4] to both sides of the equation.
  • x2 − 2x + [(b2)/4] = − 28 + [(b2)/4]
  • x2 − 2x + [(( − 2)2)/4] = − 28 + [(( − 2)2)/4]
  • x2 − 2x + [4/4] = − 28 + [4/4]
  • x2 − 2x + 1 = − 27
  • Factor the perfect square trinomial(left side) using the formula x2 + bx + [(b2)/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
x = 1 + 3i√3 and x = 1 − 3i√3
Solve by completing the square x2 − 18x + 82 = 0
  • Isolate the variables on the left side of the equation. Subtract 82 from both sides.
  • x2 − 18x = − 82
  • Complete the square by adding [(b2)/4] to both sides of the equation.
  • x2 − 18x + [(b2)/4] = − 82 + [(b2)/4]
  • x2 − 18x + [(( − 18)2)/4] = − 82 + [(( − 18)2)/4]
  • x2 − 18x + [324/4] = − 82 + [324/4]
  • x2 − 18x + 81 = − 82 + 81
  • x2 − 18x + 81 = − 1
  • Factor the perfect square trinomial(left side) using the formula x2 + bx + [(b2)/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
x = 9 + i and x = 9 − i
Solve by completing the square x2 + 12x + 1 = − 10
  • Isolate the variables on the left side of the equation. Subtract 1 from both sides.
  • x2 + 12x = − 11
  • Complete the square by adding [(b2)/4] to both sides of the equation.
  • x2 + 12x + [(b2)/4] = − 11 + [(b2)/4]
  • x2 + 12x + [((12)2)/4] = − 11 + [((12)2)/4]
  • x2 + 12x + [144/4] = − 11 + [144/4]
  • x2 + 12x + 36 = − 11 + 36
  • x2 + 12x + 81 = 25
  • Factor the perfect square trinomial(left side) using the formula x2 + bx + [(b2)/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
x = − 1 and x = − 11
Solve by completing the square x2 + 14x + 38 = − 10
  • Isolate the variables on the left side of the equation. Subtract 38 from both sides.
  • x2 + 14x = − 48
  • Complete the square by adding [(b2)/4] to both sides of the equation.
  • x2 + 14x + [(b2)/4] = − 48 + [(b2)/4]
  • x2 + 14x + [((14)2)/4] = − 48 + [((14)2)/4]
  • x2 + 14x + [196/4] = − 48 + [196/4]
  • x2 + 14x + 49 = − 48 + 49
  • x2 + 14x + 49 = 1
  • Factor the perfect square trinomial(left side) using the formula x2 + bx + [(b2)/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
x = − 6 and x = − 8
Solve by completing the square 2x2 + 4x − 8 = − 2
  • Isolate the variables on the left side of the equation. Add 8 to both sides.
  • 2x2 + 4x = 6
  • In order to complete the square, the coefficient of x must equal 1.
  • Divide everything by 2
  • [(2x2)/2] + [4x/2] = [6/2] = x2 + 2x = 3
  • Complete the square by adding [(b2)/4] to both sides of the equation.
  • x2 + 2x + [(b2)/4] = 3 + [(b2)/4]
  • x2 + 2x + [((2)2)/4] = 3 + [((2)2)/4]
  • x2 + 2x + [4/4] = 3 + [4/4]
  • x2 + 2x + 1 = 3 + 1
  • x2 + 2x + 1 = 4
  • Factor the perfect square trinomial(left side) using the formula x2 + bx + [(b2)/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
x = 1 and x = − 3

*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

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 x2 - 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 x2 = 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 22 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 ax2 + 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 b2 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 b2/4 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 b2/4? 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 b2/4.0288

But it is the same thing either way.0299

So, for example, let's say I am given x2 - 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 b2/4.0327

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

Here, a equals 1; b equals -8; therefore, I have x2 - 8x + -82/4.0339

x2 - 8x + 8 times 8...that is 64...over 4 gives me x2 - 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 b2/4, 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 x2 + 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 out0450

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 x2 + 6x = -8.0463

Now, I want to complete the square, and I know that, to complete the square, I am going to need x2 + 6x;0476

but I need a constant term, and the constant term is going to be b2/4.0483

Now, I have to add to both sides; add b2/4 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 x2 + 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, x2 + 6x + 9 = -8 + 9; or x2 + 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 b2/4 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 x2--the leading coefficient--is not 1.0658

So far, we have been working with situations where we just had x2.0664

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 coefficient0671

in front of the x2 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 2x2 - 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 x2 - 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: x2 - 6x...I need to now add b2/4 to both sides to complete the square.0730

This is x2 - 6x + b2/4 equals -2 + b2/4.0743

OK, x2 - 6x...well, b is -6, so that is + (-6)2/4...= -2 + (-6)2/4.0755

This gives me x2 - 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 x2 - 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 x2 + 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 b2/4 to both sides of this equation.0899

Well, b is 2, so that is going to give me 22/4 = -10 + 22/4.0910

So, this is x2 + 2x...well, 2 times 2 is 4; divided by 4--that is just 1.0925

So, this is -10 + 1; this gives me x2 + 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 x2 - 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 b2/4, so that is x2 - 10x + b2/4.1169

And since this is in standard form, that is ax2 + bx + c.1183

That means that b equals -10; so x2 - 10x + (-10)2, all divided by 4.1189

OK, so it is x2 - 10x +...-10 squared is 100, divided by 4; x2 - 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 b2/4.1223

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 b2/4 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 x2 + 6x + 62/4 = -12 + 62/4, 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 b2/4 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 x2 term is not 1.1451

So, if the coefficient of x2 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 x2 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 b2/4 to both sides: b is -4, so this gives me x2 - 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 x2 - 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