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Can somebody help me solve this equation?

Can somebody help me solve this equation? Topic: Can somebody help me solve this equation?
July 18, 2019 / By Aspen
Question: Solve each equation to find the exact solution, both real and imaginary. Be sure to check for extraneous roots. 1. x to the second power +6x=11 2. 4/ x-3 minus 4/x= 1 3. x to the 4th power + x to the second = 20 4. square root of 2x-5 =2 square root x-2 and last one 9x to the negative 4th power -10x to the negative 2nd power + 1 = 0 I know it's a bit difficult to do this over computer, but any help would be great! I just can't seem to figure those out! Thanks! By the way, this isn't homework, just a review sheet for a test i have tonight! So, if someone could point me in the right direction, you don't have to solve it, but just help me out that would be great! By the way, this isn't homework, just a review sheet for a test i have tonight! So, if someone could point me in the right direction, you don't have to solve it, but just help me out that would be great! Oh, and I have the answers, I just don't know how my teacher got them.
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Best Answers: Can somebody help me solve this equation?

Aaren Aaren | 10 days ago
1. x^2 + 6x = 11 (x + 3)^2 - 20 = 0 x = -3 - 2√5 x = 2√5 - 3 2. 4/(x-3) minus 4/x = 1 4x - 4(x - 3) = x(x - 3) 4x - 4x + 12 = x^2 - 3x x^2 - 3x - 12 = 0 (x - 3/2)^2 - 57/4 = 0 x = 1/2(3 - √57) x = 1/2(3 + √57) 3. x to the 4th power + x to the second = 20 x^4 + x^2 - 20 = 0 (x - 2)(x + 2)(x^2 + 5) = 0 [x^2 + (1/2)]^2 - 81/4 = 0 x = -2 x = 2 4. square root of (2x - 5) =2 square root (x - 2) √(2x - 5) = 2√(x - 2) squaring both sides: (2x - 5) = 4(x - 2) 2x = 3 x = 3/2 5. 9x to the negative 4th power -10x to the negative 2nd power + 1 = 0 9x^-4 - 10x^-2 + 1 = 0 (x - 3)(x + 3)(x - 1)(x + 1) = 0 x = -3 x = -1 x = 1 x = 3
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Aaren Originally Answered: Graphing equation question? how would solve this linear equation?
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)
Aaren Originally Answered: Graphing equation question? how would solve this linear equation?
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

Solomon Solomon
properly, the 4 could be written as 24/6, which grants an liquid crystal show and helps you to characteristic it to (x-26)/(6). so which you're left with x/5 = (24 + x - 26)/6. 24 - 26 is -2, so which you have x/5 = (-2 + x)/6. Multiply the two factors via 6 and you have 6x/5 = -2 + x. multiply the two factors via 5 and you have 6x = -10 + 5x (do not forget to distribute the 5 to the two the -2 and the x). Now, subtract 5x from the two factors and you're left with x = -10.
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Osborn Osborn
1) 54.3 2) x = 5 3) not too sure 4) What are you feeding these questions 5) Oh come on that was offside
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Osborn Originally Answered: Solve this equation; like a quadratic equation.?
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

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