Solve each equation using the quadratic formula. See Examples ...

Question

Solve each equation using the quadratic formula. See Examples 5 and 6.

$(4 x - 1) (x + 2) = 4 x$

Accepted Answer

Hello, everyone. We're going to use the quadratic formula to solve the equation 16 times eight X minus one times X plus one equals negative 15. If you recall, when we use the quadratic formula, we want our equation to be set up in the format A X squared plus BX plus C equals zero. Looking at my equation, it is not currently set up that way. So I need to do all the multiplication that's on the left hand side and then make it equal to zero. So starting out, I'm going to rewrite the 16 and then I'm going to multiply the eight X minus one times the X plus one. So I use foil for that and I get eight X squared plus eight X minus one, X minus one In my parentheses. And it still equals negative 15. Next, I'm distributing my 16th into the parentheses. 16 times eight X squared gives me 128 X squared plus 128 X minus 16, X minus 16 equals negative 15. I'm gonna come over here and combine my like terms. So I have 128 X squared plus X minus 16 equals negative 15. And then to make this equal to zero, I add 15 to both sides. And the equation I'm going to be working with is 128 X squared plus 112 X minus one equals zero. So now that I know this, I can assign my A B and C because now it matches the format. So from our equation a is gonna be B is 112 And C is -1. Recall that the quadratic formula is that X equals the opposite of B plus or minus the square root of B squared minus four A C all over two A. So now that we have it in the right format, we've identified what A B and C are. I'm gonna plug it into that equation. So I'm gonna do so right here. So X equals the opposite of beast. The opposite of negative 1 12, which is I'm sorry, the opposite of positive 1 12, which is negative 1 12 plus or minus the square root of 112 squared minus four times A which is 128 times C which is negative one, All of that over two times a. So two times 128. So next, I'm going to do some of the multiplication that's under the radical. The negative 1 20 I'm sorry, the negative 1 12 stays on the outside plus or minus. And then the square root 112 squared is and then negative four times 128 times a negative one Is Plus 512. And all that's over two times 1 28, which is 256. We have some big numbers but we'll get them down a little bit Next, I'm adding together what's under my radical. So I have negative 112 still on the outside plus or minus the square root of when I add those together, I get 13, Over 256. So I don't think 13,056 is a perfect square. Um But Nice little helpful hint is I'm looking at my answer choices and noticing that the denominator in all of them is 16, which makes me think this is reducible. So I'm gonna reduce My square root of 13,056. I'm gonna see what I can reduce that to and then I'll plug it back into the quadratic formula. So looks like a perfect square that goes into that is 256 And then times the square root of 51. So square to 256 is Times the square to 51. So when we reduced the square root of 13,056 it reduces to 16 radical 51. So I'm gonna plug that back into my formula. So I have negative 112 plus or minus 16 radical 51 Over 256. And I checked quickly in a calculator, I can reduce 112 x 16. I can also reduce 256 x 16. So I'm gonna divide 16 out of each term there. And when I do that, I get that it's negative seven plus or minus the square to 51 Over 16, which if I want to write as two different x values is negative seven Plus the square to 51/16 and negative seven minus the square root of 51/16. Writing it as a solution set, I should put the minus before the plus. Um So checking that in our answers, it looks like that is answer choice. A have a nice day.

Solve each equation using the quadratic formula. See Examples 5 and 6.
$(4 x - 1) (x + 2) = 4 x$

Key Concepts

Quadratic Equations

Expanding and Simplifying Expressions

Quadratic Formula

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