WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In previous lessons, we talked about solving quadratic equations through factoring and through graphing.
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Today, we are going to learn another method called completing the square.
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And you may have learned this earlier on in Algebra I, so it may be review.
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And if you need even more depth, you can look at the Algebra I series here at Educator.com.
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Now, recall that some quadratic equations can be solved by taking the square root of both sides of the equation.
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And this includes equations where the quadratic expression is a perfect square.
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So, for example, if you have x² - 8x + 16 = 5, you may recognize that this is a perfect square.
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This trinomial is a perfect square trinomial.
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It is actually equal to (x - 4)², or (x - 4) (x - 4).
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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.
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So, take the square root of both sides of the equation to get x - 4 = ±√-5.
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And recall that you need to take both the positive and negative square root of 5,
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because, thinking back, if you had something such as x² = 4,
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and you were to take the square root of both sides, that would give you that x could equal 2,
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or x could equal -2, because 2² equals 4, and (-2)² equals 4.
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So, it is the same concept here; it looks more complex, but it is the same concept.
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On the left, I am taking the square root, and then on the right, I am taking the square root of a number;
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and it could be both positive square root of 5, or the negative whatever-the-square-root-of-5-turns-out-to-be.
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And I can't simplify that any further without a calculator; I can just leave it in this form, ±√5.
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Since this is an irrational number, I can't get an exact value; I just need to leave it in this format.
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So, this is going to give me x = 4 ± √5.
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And I could put this in this form; I could leave it; or I could say x = 4 + √5, and x = 4 - √5.
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These are both solutions to this quadratic equation.
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Now, how do you recognize perfect squares?
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If you have a perfect square trinomial, and it is in the form ax² + bx + c,
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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.
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So, if you are looking, and you are not sure if it is a perfect square trinomial, look at the constant term.
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And here, I have a = 1 for a coefficient; b = -8; and c = 16.
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So, what I want to do is see if b²/4 equals 16.
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And if I look, I see this (-8)²/4, and that equals 64 divided by 4, which is 16.
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So, this checks out; so if you are not sure if you have a perfect square trinomial, you can just look at the constant.
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If the rest of it checks out as perfect squares, and then you are not sure,
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then you can check and say, "OK, does the constant equal b²/4? Yes, it does."
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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.
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OK, sometimes you don't have a perfect square trinomial; in that case, what you can do is actually complete the square.
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You can make it into a perfect square.
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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.
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But it is the same thing either way.
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So, for example, let's say I am given x² - 8x, similar to the last one (but that one was already a perfect square).
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So, I am just given this portion; and I want to complete the square.
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So, in order to complete the square, what I need is a constant term.
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And I know that, in a perfect square trinomial, the constant is going to equal b²/4.
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So, really what I am looking for is this.
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Here, a equals 1; b equals -8; therefore, I have x² - 8x + -8²/4.
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x² - 8x + 8 times 8...that is 64...over 4 gives me x² - 8x + 16.
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And now, I have a perfect square trinomial.
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So, last time I talked about if you had something and you thought it was a perfect square trinomial;
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you could check it out by seeing if the constant is equal to this.
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Conversely, if you don't have a perfect square, and you need to complete the square--
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you just have this part, and you want to get the constant term to complete the square--
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you can use this knowledge to find what that constant should be; and then, from there, we can solve equations.
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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.
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In the last slide, I talked about completing the square; but if you have an equation, you need to keep the equation balanced.
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So, you can't just complete the square, and then do nothing to the other side.
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So, let's take an example here, slightly different than the other one, but related.
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This time, I have x² + 6x + 8 = 0; so now I have an equation.
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Now, the first thing to do is: if you are dealing with an equation, get the constants on one side.
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I am going to get the constants on the right, because I want to clear out
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and just have these x variable terms here on the left, so I can complete the square on that.
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I am going to subtract 8 from both sides to get x² + 6x = -8.
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Now, I want to complete the square, and I know that, to complete the square, I am going to need x² + 6x;
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but I need a constant term, and the constant term is going to be b²/4.
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Now, I have to add to both sides; add b²/4 to both sides.
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This is the step that sometimes gets forgotten, and then you get an incorrect answer.
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OK, this is going to give me x² + 6x +...well, b is 6; now, I have to add that over here, as well; OK.
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This is going to give me 36 divided by 4, which comes out to 9.
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So, coming up here, x² + 6x + 9 = -8 + 9; or x² + 6x + 9 = 1.
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Now, I have a perfect square trinomial, and I can go and use techniques used previously in order to solve this--
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such as taking the square root of both sides.
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(x + 3)² (which is the same as this trinomial) equals 1.
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Now, we take the square root of both sides to solve this to get x + 3 = ±√1.
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So, the square root of 1 is 1, so I get x + 3 = ±1.
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From there, you can just solve; I get x + 3 = 1, so x = -3 + 1; x = -2;
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and I also have x + 3 = -1; therefore, I am going to get x = -4.
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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.
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To solve a quadratic equation when you don't already have a perfect square trinomial involved,
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get the constants on the right and the variables on the left.
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Then, complete the square by adding b²/4 to both sides.
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Once you have done that, you have a perfect square here on the left.
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Turn this into this form, and then take the square root of both sides to get your solutions.
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Sometimes, the coefficient of x²--the leading coefficient--is not 1.
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So far, we have been working with situations where we just had x².
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If the coefficient is not 1, you need to take an extra step,
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and you need to divide both sides of the equation by the coefficient
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in front of the x² term in order to make that coefficient 1.
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Then, you just go on and complete the square as usual.
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For example, if I was given 2x² - 12x + 4 = 0, I need to make this coefficient 1.
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So, I can do that by dividing both sides by 2.
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And that is going to cancel out right here to give me x² - 6x + 2 = 0.
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OK, now I am going to go ahead and complete the square.
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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.
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Then, I need to complete the square: x² - 6x...I need to now add b²/4 to both sides to complete the square.
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This is x² - 6x + b²/4 equals -2 + b²/4.
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OK, x² - 6x...well, b is -6, so that is + (-6)²/4...= -2 + (-6)²/4.
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This gives me x² - 6x; this is 36 divided by 4; and I am just going to simplify that to 9.
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And if I had not added to both sides, the equation would not be balanced; I would not get the correct solutions.
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OK, I am coming up here to get x² - 6x + 9 = 7.
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Well, now I have a perfect square trinomial, which is actually (x - 3) (this is a negative right here) squared equals 7.
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Remember to take the square root of both sides to get x - 3 = ±√7.
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Isolate the x to get x = 3 ± √7.
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So, the two solutions are x = 3 + √7 and x = 3 - √7.
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OK, so it is the same as what we just did with completing the square, and then taking the square root of both sides;
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but with an extra step, because in order to complete the square, you need for the coefficient x to be 1.
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If it is not 1, if you have a leading coefficient that is not 1, begin by dividing both sides by that coefficient.
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OK, sometimes the solutions to a quadratic equation may be complex numbers.
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Just to review: complex numbers consist of a real part and an imaginary part, something such as 3 + 2i.
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So, this is in the form a + bi, where a is the real part, and 2i is the imaginary part.
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It turns out that, for quadratic equations, solutions may end up in this form.
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Let's look at a situation where that could occur.
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Let's say I am given x² + 2x + 10 = 0.
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And I am going to solve this by completing the square.
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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.
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Now, I need to complete the square.
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I want to add b²/4 to both sides of this equation.
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Well, b is 2, so that is going to give me 2²/4 = -10 + 2²/4.
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So, this is x² + 2x...well, 2 times 2 is 4; divided by 4--that is just 1.
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So, this is -10 + 1; this gives me x² + 2x + 1 = -9.
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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.
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So, let's come up here to finish this and take the square root of both sides, and we rewrite it.
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This is a perfect square: it is (x + 1)² = -9.
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The square root of both sides gives me x + 1 = √-9.
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Now, before learning about complex numbers, we would have had to stop here and say,
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"OK, this is undefined; there is no real number solution; we don't know what to do with this."
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But now that we have discussed imaginary numbers, we do know what to do with this.
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So, let's look at how we can handle this.
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We know that this equals √-1 times 9; and remember the positive and negative results for this: you need to take both: ±√-9.
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OK, using the product property, this is going to equal this.
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Recall that, with complex numbers, the square root of -1 is equal to i.
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OK, so this is going to give me x + 1 = ± i times √9, which is 3.
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Therefore, now this is something I can work with: x = (I am going to subtract -1) ± 3i.
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All right, so here we have a situation where the solution to the quadratic equation is a complex number.
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And the reason is: you are taking the square roots.
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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.
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So here, I ended up getting all the way down by my usual method, completing the square,
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then taking the square root of both sides; and I ended up with this.
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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.
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And that is a set of complex conjugates.
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So, let's use the square root property to solve this.
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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.
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Since this is negative here, this is (x - 3)² = 7.
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And if I take the square root of both sides, I am going to get x - 3 = ±√7, so x = 3 ± √7.
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Or, you could write this out as x = 3 + √7, and x = 3 - √7, as the solutions.
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When it is already a perfect square trinomial, it is much easier.
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In this example, I am going to have to complete the square.
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And this is just asking me to complete the square of x² - 10x.
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So, in order to do that, I need a constant.
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And for a perfect square trinomial, the constant is going to equal b²/4, so that is x² - 10x + b²/4.
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And since this is in standard form, that is ax² + bx + c.
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That means that b equals -10; so x² - 10x + (-10)², all divided by 4.
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OK, so it is x² - 10x +...-10 squared is 100, divided by 4; x² - 10x + 25.
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This is a perfect square; this is the same as (x - 5)².
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So, to complete the square, you need to find a constant term; and the constant term is going to be equal to b²/4.
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So, b equals -10; substituting in here gave me a perfect square trinomial.
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Solve by completing the square: the first step is to isolate the x variables on the left side of the equation.
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So, subtract 12 from both sides.
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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.
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OK, b is 6, so this gives me x² + 6x + 6²/4 = -12 + 6²/4, which is 36/4; 36/4 is just 9.
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We will find a bit more on the right; now I have my perfect square trinomial on the left.
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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.
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Now, I am going to write this as (x + 3)² = -3.
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In order to solve this, I am going to take the square root of both sides.
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x + 3 = ±√-3; from previous lessons, I know that I can change this to -1 times the square root of 3.
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And recall that i is equal to the square root of -1, so this is i√3.
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OK, this gives me x = -3 ± i√3; or I can write it out as x = -3 + i√3, and x =....
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These are both solutions; they are not real-number solutions, but they are solutions.
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Solving by completing the square means isolating the x variable terms on the left and the constants on the right.
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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.
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Solving by taking the square root of both sides gave me this.
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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.
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Again, we are going to solve by completing the square.
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And my first step is always to get the variable terms on the left and the constants on the right, then complete the square.
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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.
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So, if the coefficient of x² is not 1, divide both sides of the equation...
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And the reason is that I need to do that in order to complete the square and go on as usual.
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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.
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These cancel out, and now my x² coefficient is 1.
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12 divided by 3 gives me -4x; and 6 divided by 3 is 2.
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Now, I can go about completing the square, because my leading coefficient is 1.
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I am doing that by adding b²/4 to both sides: b is -4, so this gives me x² - 4x, and this is...
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4 times 4 is 16, divided by 4; so that is just 4; and this is also going to be 4.
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OK, coming up here to finish, this is going to give me x² - 4x + 4 = 6.
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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.
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And I am going to find the square root of both sides; so x equals 2 plus or minus the square root of 6.
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This problem took an extra step: after isolating my variables on the left and my constants on the right,
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I saw that my leading coefficient was not 1, so I had to divide both sides of the equation by that coefficient 3.
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Once I was there, then I went about completing the square by adding this term to both sides,
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getting a perfect square trinomial (which is this), and taking the square root of both sides to get my solutions.
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Thanks for visiting Educator.com; and I will see you for the next lesson.