Solve each equation in Exercises 47–64 by completing the squar...

Question

Solve each equation in Exercises 47–64 by completing the square. $2 x^{2} - 7 x + 3 = 0$

Accepted Answer

Welcome back. I'm so glad you're here. We've got this equation four X squared minus X plus 15 equals zero. And we are asked to solve this by completing the square. Well I'll go through the steps of completing the square but if this is a little bit fuzzy, definitely go back and watch some previous lesson videos on completing the square. Just for a refresher. I want to start by pointing out that this is an equation in the form of A X squared plus B, X plus C equals zero. So I'm going to be referring to our A B and C terms as we go throughout our A term is four B term is negative 23 are negative 23. Excuse me and r C term is 15. We want to start off completing the square by ensuring that are a term equals one. So we're going to go through and we're going to divide everything by a when that a term starts off as being one in the first place, it's much simpler. So doing that division by four we now have X squared minus 23 4th X plus 15 4th equals while still zero. And the next thing we want to do is we want to bring our C term or in this case RC over a term because a wasn't equal to one over to the right hand side of the equation. So in this one in black we're subtracting 15/4 from both sides. So that gives us X squared minus 23 4th X equals negative 15 4th and where you had that C term or that C over a term. I want you to leave a space because we are going to go back and fill in that space carrying down the red. We've got X squared plus B over A. X. Space equals a negative C over a space. So how do we fill in those spaces? What we're going to do is we're going to take our B term or B over A. When A wasn't equal to one. We're going to divide that by two and we are going to square it. So what is our B term in the black? That is negative 23 4th We're going to take negative 23 4th divided by two and square it. I'll rewrite this as negative 23/4 divided by two squared. Which reminds me it's the same thing as negative 23 4th times one half squared. And when we do that multiplication we get in the numerator negative 23 in the denominator eight and we'll square that. But before we square it. I want you to pause when you've got something still in the parentheses that needs to be squared. I want you to take note of what that number in the parentheses is. Or in other words that be over a over two and hold on to that number for a moment. And then yes we're going to continue on and square it. So that equals over 64 and this is our number that's going to get filled in. Um two are blank spaces. So let's go ahead and do that. We've got x squared minus 23 4th X plus 64th equals negative 15 4th plus 64th. Really important to add that to both sides and in red that will be X squared plus B over A X plus B over a over two squared equals a negative C over A plus RB over a over two squared. Okay, now we need to recall something about completing the square. That by its very definition when you have X squared plus B over A X plus B over a over two X. Or squared. Excuse me. That is the same thing as parentheses X Plus the number we highlighted. R. B over A over to end parentheses squared. It's not that whole thing. Just this second one highlighted. So that's why we had to hold on to that number. So instead of X squared minus 23 4th X plus 529 64th that is parentheses X plus are highlighted number negative 8th and parentheses squared equals and you can bring down the right hand side negative 15 we can combine that negative. 15 4th plus 529 64th is 289 64th. Wonderful. Okay, the next step we're going to do now is basically just simplifying this to solve for X. The factor on the left hand side is squared so we can go ahead and take the square root of both sides. On the left hand side, the square root and the squared cancel out. So we've just got X plus a negative 23 8th equals. And on the right hand sign That'll be a plus or minus because we're taking the square root, square root of 289 is 17 and the square root of 64 is eight. Now to get X by itself, all we got to do is add 23/8 to both sides. We're almost there notice that we have a plus or minus. We actually have two equations here we have X equals Are positive 17/8 plus 23/8 and we have our X equals the negative 17 8th plus 8th. And going ahead and doing that addition of our fractions, we get X equals five for one of our solutions. And doing the addition of the negative value negative 17 8th plus 23 8th. We get X equals 3/4 for our other solution. So I'll write that up here so that we can still see it when we scroll back up to the top, we look at our answer choices and this matches with answer choice. C Well done. We'll catch you on the next one

Solve each equation in Exercises 47–64 by completing the square. $2 x^{2} - 7 x + 3 = 0$

Key Concepts

Completing the Square

Quadratic Equation Standard Form

Isolating the Variable

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Solve each equation in Exercises 47–64 by completing the square. 2x2−7x+3=02x^2 - 7x + 3 = 0

Key Concepts

Completing the Square

Quadratic Equation Standard Form

Isolating the Variable

Watch next

Solve each equation in Exercises 47–64 by completing the square. $2 x^{2} - 7 x + 3 = 0$