Solve each equation using the quadratic formula. See Examples 5 a...

Question

Solve each equation using the quadratic formula. See Examples 5 and 6. 2/3x^2 + 1/4x = 3

Accepted Answer

Hello, everyone. We're gonna solve the equation X squared plus 14 8th, X minus 147/16 equals zero using the quadratic formula. So I want to rewrite this because I don't like seeing my fractions horizontally. I'd rather see them vertically. And when I do, I have X squared plus 14/8 times X minus 147/16 equals zero. Now, I want to use the quadratic formula and this is set up in the right format, but I really don't want to deal with those fractions. There is a way I can get rid of those fractions. If you recall, if I multiply the entire equation through by a common denominator, I can cancel out the denominators. And that way I can get rid of some of the fractions. So our common denominator here would be 16. And when I do that 16 times X squared gives me 16 X squared, 16 times 14/8 X, it would cross reduce, leaving me with two times 14. So this would be plus 28 X third term. When I cross reduced there, the sixteens would cancel out. So I have minus 147 and 16 times zero is zero. I can do this without changing any of my values. And this is probably going to be easier to plug into our equation. So recall that when we're working with the quadratic formula, we are looking at a format that is A X squared plus BX plus C equals zero. So now that I have it in that format, I'm gonna identify my A B and C A 16 B is 28 and C is negative 147. Also recall that the entire formula which is X equals negative B plus or minus the square root of B squared minus four AC over two A. So now that I've identified my A B and C, I'm gonna plug it into the formula, we're gonna end up with some larger numbers, but we can break them down pretty quickly. So X equals the opposite of B. So negative 28 plus or minus the square root of 28 squared minus four times A which is 16 times C which is negative 147 all of that over two times A. So two times 16, I'm gonna multiply through what's under my radical outside. I still have negative 100 tw oh sorry, negative 28 I made it bigger than it needed to be inside the square root or squaring 28 gives me 784 negative four times 16 times negative 147 multiplies to a positive 9408. And then in the denominator two times 16 is 32. Adding together what's under the radical? The negative 28 still on the outside plus or minus the square root of when I add those together, I get 10,192 over 32. Now, I definitely want to reduce the square root of 10,192. I will tell you that off the top of my head. I do not know the perfect square there. I don't know what the largest perfect square is that goes into it. I kind of have to guess and check what I am gonna use as a hint though. Is that in my answer, we have radical 13. So I'm gonna say that radical 13 is one of the things I'm gonna use to reduce it. And when I do that, this is the squared of 784 times a square to 13 that equals 10,192 square 100 of seven square root. I'm sorry of 784 is 28. So we can reduce this radical to 28 times the square to 13. So I'm gonna plug that back into my formula. So now I have negative 28 plus or minus 28 radical 13 over 32. I want to reduce those fractions. Um All the terms look like they're divisible by four. So I'm gonna do that. And when I do that, I get negative seven plus or minus seven radical 13 over eight. In writing that as a solution set, I have negative seven minus seven radical 13/8 and negative seven plus seven radical 13 over eight. Those are my two solutions. And when I check over with my choices that looks like it is answer choice B have a nice day.

Solve each equation using the quadratic formula. See Examples 5 and 6. 2/3x^2 + 1/4x = 3

Key Concepts

Quadratic Formula

Standard Form of a Quadratic Equation

Discriminant

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