Question

(x+3)^1/2-1=x
Please help me solve that

Answer 1

you need help still

Answer 2

Find the partial-fraction decomposition of the following expression:
(x^2 + 1) / [ x (x - 1)^3 ]

The factor x – 1 occurs three times in the denominator. I will account for that by forming fractions containing increasing powers of this factor in the denominator, like this:

(x^2 + 1)/[x(x - 1)^3] = A/(x - 1) + B/(x - 1)^2 + C/(x - 1)^3 + D/x

Now I multiply through by the common denominator to get:

x2 + 1 = Ax(x – 1)2 + Bx(x – 1) + Cx + D(x – 1)3

I could use a system of equations to solve for A, B, C, and D, but the other method seemed easier. The two zeroing numbers are x = 1 and x = 0: so

x = 1: 1 + 1 = 0 + 0 + C + 0, so C = 2
x = 0: 1 = 0 + 0 + 0 – D, so D = –1

But what do I do now? I have two other variables, namely A and B, for which I need values. But since I've got values for C and D, I can pick any two other x-values, plug them in, and get a system of equations that I can solve for A and B. The particular x-values I choose aren't important, so I'll pick smallish ones:

x = 2: Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved
(2)2 + 1 = A(2)(2 – 1)2 + B(2)(2 – 1) + (2)(2) + (–1)(2 – 1)3
4 + 1 = 2A + 2B + 4 – 1
5 = 2A + 2B + 3
1 = A + B

x = –1:

(–1)2 + 1 = A(–1)(–1 – 1)2 + B(–1)(–1 – 1) + (2)(–1) + (–1)(–1 – 1)3
1 + 1 = –4A + 2B – 2 + 8
2 = –4A + 2B + 6
2A – B = 2

I'm still stuck solving a system of equations, but by using the easier method to solve for C and D, I now have a simpler system to solve. Adding the two equations, I get 3A = 3, so A = 1. Then B = 0 (so that term in the expansion "vanishes"), and the complete decomposition is:

1/(x - 1) + 2/(x - 1)^3 - 1/x

Answer 3

yes - as simple as that

Answer 4

Oh man…. that's confusing.. sorry @ayla2001 but would you be able to just work with me step by step?

Answer 5

x=1
x in (-oo:+oo)

((x+3)^1)/2-1 = x // - x

((x+3)^1)/2-x-1 = 0

(x+3)/2-x-1 = 0

(x+3)/2+(2*(-x))/2+(-1*2)/2 = 0

x+2*(-x)-1*2+3 = 0

3-x-2 = 0

1-x = 0

(1-x)/2 = 0

(1-x)/2 = 0 // * 2

1-x = 0

1-x = 0 // - 1

-x = -1 // * -1

x = 1

Answer 6

(x + 3)^1/2 - 1 = x
add 1 to both sides
(x + 3)^1/2 = x + 1
square both sides
x + 3 = (x + 1)^2
x + 3 = x^2 + 2x + 1
x^2 + x - 2 = 0

now solve this for x

Answer 7

do you get it know

Answer 8

Got it thanks. Would yo be able to help with another problem?

Answer 9

the way you wrote the problem is ambiguous - it can be interpreted in different ways