Solve each equation using completing the square. See Examples 3 a...

Question

Solve each equation using completing the square. See Examples 3 and 4. -3x^2 + 6x + 5 = 0

Accepted Answer

Welcome back. I am so glad you're here. We're told in the given equation, complete the square and solve it. Our given equation is 121 X squared minus 242 X plus 101 equals zero. And our answer choices are answer choice. A the solution set the quantity of one minus two square at five all divided by 11. And the quantity of negative one plus two square at five all divided by 11 answer choice B the solution set the quantity of negative one minus two square at five all divided by 11. And the quantity of one plus two square at five all divided by 11. Answer choice C the solution set the quantity of negative 11 minus two square at five all divided by 11. And the quantity of negative 11 plus two square at five all divided by 11. And answer choice D the solution set the quantity of 11 minus two square at five, all divided by 11. And the quantity of 11 plus two square at five all divided by 11. All right. When we are completing the square, the first thing we need to do is make sure that the coefficient of our X squared is one. And in this case, it is not, it's 100 21. So we need to divide all of our terms by 121 so that the coefficient of our X squared will be one. So let's do that. Division 121 X squared divided by 121 is just X squared. Yay negative 242 X divided by 121 is a negative two X positive 101 divided by 121. I'm going to leave that as that fraction. So plus 101 firsts and that's still equal to zero because zero divided by 121 is zero. All right, we've got the coefficient of our X squared as one great. The next thing we want to do is move our constant term, which is 101 120 firsts over to the right hand side of the equation. And to do that because it's positive, we will be subtracting 101 120 firsts from both sides. And then we're going to leave a big space where the constant term was. So we'll have X squared minus two X and then that cancels out, but we'll leave a big space there and it equals on the right hand side, zero minus 101 121st is a negative 101 120 firsts. And we leave a space after that term on the right hand side as well. And now we'll be filling in that space and completing the square, we fill it in by taking the coefficient of our X term. So that's negative two, we're going to divide it by two and then square it. So dividing negative two by two will be a negative one. We still need to square that but do take note for a moment of this number inside of the parentheses, the negative one, we'll need that again in a minute. Negative one squared is a positive one. So we will be adding one to both sides of the equation in those spaces. And now we can factor the left hand side, we have X squared minus two X plus one. And we know because of the definition of completing the square that, that factors to be X. And then this term here that we have inside of the parentheses before we squared it, it'll be X minus one and that quantity is squared. And that's our left hand side. Now, on the right, we have a negative 101 120 firsts plus one. We are adding and we have a fraction, we need to have a common denominator that's going to be 121. So we'll multiply one by 121 firsts or in other words, we're adding negative 101 firsts plus 120 firsts. And now we have a common denominator. So we can add straight across in our numerators. Negative 101 plus 121 is a positive 20. Our denominator remains 121. So our quantity of X minus one squared is equal to 2120 firsts. Now, we're still trying to solve for X. We want to get X by itself right now. It's being squared, we need to take the square rut and what we do to one side, we do to the other as well. On the left hand side, taking the square rut of the quantity of X minus one squared is just X minus one. That side is looking pretty good on the right. We recall from previous lessons that when we take the square rut of a fraction, we do need to note this will be a plus minus now that we're taking it and we can take the square root of both the numerator and the denominator. So we have the square root of 20 over the square root of 121. And as far as the denominator is concerned, we have plus or minus the denominator, the square root of 121 that's a positive 11 in the numerator, the square root of 20. Well, we can simplify that a little bit. We know that if we take the square rut of 20 that's the same as the square rut of four multiplied by five, which is the same as the square, it of four multiplied by the square rot of five. The square of four is two. So we have two square rot five in our simplified numerator. So X minus one is equal to plus or minus two squarer five 11th. Now X is almost by itself, let's add one to both sides of the equation because on the left, that negative one plus one, that's zero. And all we've got is X yay, the X is happy. It's by itself. And then on the right, we have a positive one plus or minus two square at 5/11. Now we can combine the one and the two square root 5/11. We're adding with fractions need a common denominator which will be 11. We can multiply the one by 11 11th. So we'll have, oops wanted to switch colors. So we'll have 11 11th plus or minus two square 5/11. We have a common denominator. So we can add our numerators which will just be plus two square at five 11th and 11 minus two square up five 11th. And those are our two answers for X. So we look at our answer choices and that will match answer choice. D well done. We'll catch you on the next one.

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Solve each equation using completing the square. See Examples 3 and 4. -3x^2 + 6x + 5 = 0

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