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

Question

Solve each equation using the quadratic formula. See Examples 5 and 6. 1/2x^2 + 1/4x - 3 = 0

Accepted Answer

Hello, everyone. We're going to use the quadratic formula to solve the equation one half X squared plus 1/4 X minus 25/ equals zero. So first, I'm going to rewrite this. So my fractions are vertical. So one half times X squared times 1/4 times X minus 25/8 equals zero. I recall to use the quadratic formula. I want my equation as A X squared plus BX plus C equals zero. So we're almost there, but I don't want to leave these fractions. That's just gonna make things so much harder than it needs to be. I recall that I can multiply an equation through by a common denominator and that will cancel out all of the fraction pieces. So in this case, our common denominator would be eight. So I'm gonna distribute eight into this and this is not an exponent. I'm just multiplying. So eight times one half is four. So I get four X squared, eight times 1/4 is two. So two X and then eight times negative 25 8th, the eighth will cancel out and I'll be left with 25. So now I have four X squared plus two, X minus 25 equals zero. And that's gonna be a lot easier to work with. So I'm gonna identify my A B and C based on the base equation. So a is four, B is two C is -25. Recalling that the quadratic equation is that X equals negative B plus or minus the square root of B squared minus four A C all over two A. I'm now going to take the values I found for A B and C and plug them into that formula. So X equals negative B or the opposite of B. So negative two plus or minus the square root of B squared. So two squared minus four times A which happens to be four times C which is negative All over two eggs or two times 4. Now I'm gonna do the multiplication under the radical. So X equals negative two still on the outside. Lost their minus the square root of radical. Two squared is four, negative four times four times negative 25 is a positive 400 Over two times 4 which is eight. Next, I'm gonna add what's under my radical? So X equals negative two plus or minus the square root of 404 over eight. I want to reduce that radical. I'm gonna do it off to the side and then plug it back in. So the square root of 404 that is not a perfect square. But I do recognize that I could factor and get the square to four times a squared of 101. there might be something bigger. But if nothing else, I know four square 24 is two. And I don't think I can break down the square root of 101 any further. So the reduced version of my square root of 404 is too radical 101, which I'm going to put back into my quadratic formula. So I get negative two plus or -2 radical 101 over eight. Now I want to reduce that if I can. And it looks like each term is reducible by two Which gives me -1 plus your minus the square root of over four. And if I want to break that down to my solution set, I have two X values X would be negative one plus the square root of 101 over four And X would equal negative 1 - the square root of 101 over four. So those were two solutions, two possible x values. And when I check, it looks like that is answer choice a have a nice day.

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Solve each equation using the quadratic formula. See Examples 5 and 6. 1/2x^2 + 1/4x - 3 = 0

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