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Topic: Solve the equation. x^2 - x = 20?**Question:**
i can do trial and error but how do i solve this?
factor by grouping:
r^3 + r^2 + 7r + 7
perform the indicated operation:
3x^2 (-8x^3 -5x^2 + 4x - 10)

July 18, 2019 / By Luke

Trial and error, or factoring, is the fastest way to solve problem 1. 1. x² - x = 20 x² -x - 20 = 0 (x - 5)(x + 4) = 0 x = 5 OR x = -4 2. r³ + r² + 7r + 7 r²(r + 1) + 7(r + 1) (r² + 7)(r + 1) 3. 3x²(-8x³ - 5x² + 4x - 10) -24x^5 - 15x^4 + 12x³ - 30x²

👍 166 | 👎 9

Did you like the answer? 2x + 3(11/3) = 12 3/1 * 11/3 = 33/3 When multiplying, just multiply across, numerator to numerator and denominator to denominator. (I know I put a 1 under the 3, all whole numbers have a 1 under them) (example : 3/1 = 3) or ( lets say 5 in fraction form is 5/1 ) (Another fraction example 2/5 * 3/4 mult across and you get 2 * 3 = 6 and 5 * 4 = 20 answer 6/20 then, simplify and get 3/10) Now simplify 33/3 = 11 2x + 11 = 12 Now subtract 11 from both sides. and you get 2x = 1 2x = 1 Now divide 2 on both sides and you get x = 1/2 ordered pair is (1/2 , 11/3)

2x + 3(11/3) = 12 look at 3(11/3) = (3 * 11) / 3 = 11 ( 3/3) = 11 that gives you 2x + 11 = 12 2x = 1 x = 1/2 the point (1/2 , 11/3) is a solution to the equation remember, when you multiply fractions, you multiply the tops and multiply the bottoms if there isn't a bottom for one of the fractions, treat it like 1 3(11/3) = (3/1)(11/3) = (3*11) / (1 * 3) = 33 / 3 = 11

2x + 3 (11/3) = 12 first you multiply the 3 into 11/3 and you'll now have 2x + 11 = 12 next subtract 11 from both sides 2x = 1 finally divide by 2 in both sides x = 1/2 or .5

Ok move the 20 to the other side so it's x^2 - x - 20 = 0. Then 5*4=20 and 4-5=-1 (coefficient of x). So you can split it up like x^2 + 4x - 5x - 20 = 0. And factor from there so you can pull out an x from x^2 + 4x so it looks like x(x+4) <-- what's left once you pull out the x. And from -5x -20 you want to get the same thing in te parentheses so pull out a -5 and get -5(x+4). And together it's (x+4)(x-5) and solve for x. You get x= -4 and 5 and when you test it out you get 5 as the answer!

👍 60 | 👎 7

With substitution it becomes a quadratic. Let u = w² and rewrite as: u² + 7u - 18 = 0 (u+9)(u-2) = 0 u = -9 and 2 = w² w = ±3i and ±√2

Quartic equations have quadratic factors. w⁴ + 7w² - 18 = 0 can be factored as (w² - 2)(w² + 9) = 0. This means that w² - 2 = 0 or w² + 9 = 0. w² - 2 = 0 means that w² = 2, so w = ±√2. w² + 9 = 0 means that w² = -9, so w = ±√(-9) = ±3i. The solution set is {√2, -√2, 3i, -3i}.

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