Mathematics

Mathematics, 04.07.2019 15:30, asims13

Which equation has the solution x=1+- sqrt 5?

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Mathematics, 23.06.2019 07:20, ANONYMUSNESS8670

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution. u = x 4(x +1) x +1 (x + 1)2 select the solution(s) of the original equation. x = 1 + sqrt 2 x = 2 + sqrt 2 x = 3 + sqrt 2 x = 1 - sqrt 2 x = 2 - sqrt 2 x = 3 - sqrt 2

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Mathematics, 14.07.2019 19:00, laurenppylant

Solve the equation and express each solution in a+bi form. x^4-7x^2-8=0 a. x = -1, 1, 2 sqrt 2, or -2 sqrt 2 b. x = -i, i, 2 sqrt 2, or -2 sqrt 2 c. x = -1, 1, 2 sqrt 2i, or -2 sqrt 2i d. x = -i, i, 2 sqrt 2i, or -2 sqrt 2i

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Mathematics, 26.09.2019 05:30, 5001Je

2. the equation x+2= sqrt (3x+10) is of the form x+a=sqrt(bx+c) , where a, b, and c are all positive integers and b > 1. using this equation as a model, create your own equation that has extraneous solutions. (a) using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt(bx+c) where a, b, and c are all positive integers and b > 1 such that 7 is a solution. (hint: substitute 7 for x, and choose a value for a. then square both sides so you can choose a, b, and c that will make the equation true.) (b) solve the equation you created in part 2a. (c) if your solution in part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. if your solution in part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.

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Mathematics, 25.10.2019 18:43, Jana1517

The equation x+2=sqrt(3x+10) is of the form , x+a=sqrt(bx+c),where a, b, and c are all positive integers and b> 1 . using this equation as a model, create your own equation that has extraneous solutions. using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt(bx+c) , where a, b, and c are all positive integers and b> 1 such that 7 is a solution. (hint: substitute 7 for x, and choose a value for a. then square both sides so you can choose a, b, and c that will make the equation true.) solve the equation you created in part 2a. if your solution in part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. if your solution in part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.

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