Question

How do you solve this problem??
Any help is appreciated!!

4(x+1)^1/2 -5(x+1)^3/2 + (x+1)^5/2 =0

Answer 1

did you google it?

Answer 2

no..does typing it like that actually help find a way to solve it?

Answer 3

well u could type in the properties of the problem and u might get help on how to solve it like a step by step whats the unit of the subject? amd firther more what kind of math is tht put in the answers to those questions and you may find what your looking for

Answer 4

if (x+1)^1/2=y then {y^2 = x+1}
4y-5y^3+y^5=0 or y(y^4-5y^2+4)=0 y=0 is one answer,
x=-1.

y^2=z z^2-5z+4=0 delta=25-16=9
z1=(5-3)/2 =1 = y^2 = x+1 x=0

z2=(5+3)/2 =4 = y^2 = x+1 x=3

Answer 5

Thank you soo much! Do you know how to solve this one too? \[\sqrt{x} - 3\sqrt[4]{x} = 4\]

Answer 6

You are welcome!
x^1/4 =y {x=y^4}
domain of x: x>=0 {square} so y>=0 too.
y^2-3y-4=0 delta=9+16=25
y=(3-5)/2 =-1. -1>=0 is false. We can't speak about x.
y=(3+5)/2=4 4>=0 is true. so x=4^4=256

Answer 7

Once again thank you!!! do you know how to do this one?

\[\sqrt{6}x ^{2} + 2x - \sqrt{3/2} = 0\]

Answer 8

Could you please type it again?
you can show power for example x^1/2 or x^(-1/2)

Answer 9

Sure!
\[\sqrt{6}x^2 + 2x - \sqrt{3/2} = 0\]
its square root of 6 and then x^2 plus 2x minus the square root of 3/2 =0

Answer 10

it's ax^2 + bx + c =0
delta =4- 4(sqrt6)(-sqrt3/2) =4+4.sqrt(9)=4+12=16
x1=(-2-4)/2.sqrt(6) =-sqrt(6)/2
x2=(-2+4)/2.sqrt(6) = 1/sqrt(6) =sqrt(6)/6

Answer 11

Is it OK?

Answer 12

I dont really understand the third and fourth lines

Answer 13

x1=[-b-sqrt(delta)]/2a
x2={-b+sqrt(delta)]/2a

Answer 14

it's a general formula too find roots of quadratic equation.

Answer 15

If delta>0 then we have two unequal roots..
delta=0, x1=x2 . only one root.
delta<o then we have no roots in real numbers.

Answer 16

ohh ok!! thanks so much!you really dont know how much youve helped me haha.do you mind if i ask you one more?(:

Answer 17

i become glad if i could help you:)

Answer 18

\[(1/(x-1)) -( 2/x ^{2}) = 0\]

Answer 19

at first x uneqal 1 and 0 is our domain.

Answer 20

*(x-1)x^2 :
x^2-2(x-1)=0 or x^2-2x+2=0

Answer 21

delta=4-8 = -4<0

so I think we have no roots in real number.

Answer 22

you have answer?

Answer 23

no..ive been trying all these different methods and i keep on getting numbers like 2,1,-1 and i plug them back in and they dont come out to equal 0

Answer 24

I rewrite your equation: please tell me is it true?

Answer 25

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

Answer 26

excuse me minus -

Answer 27

correct. its like that on the page but to the left of the minus, its a fraction written like on top of each other and same thing with the 2/(over) x^2

Answer 28

1 / [1/(x-1)]
is it tru?

Answer 29

and -2/[1/x^2]

Answer 30

like 1/x-1 and -2/x^2 = 0
if that makes sense

Answer 31

now I see correct form of your equivalent.
thank you.
I think my answer is true.

Answer 32

so its an extraneous solution?

Answer 33

it has imaginary roots.
if our domain contain imaginary numbers: a+ib a,b belong to R.
i = (-1)^1/2

Answer 34

I think here our domain is R; Real numbers

Answer 35

ok..i havent learned about that ha

Answer 36

imaginaru nums? no problem. not impportant now.
but solution in real numbers>

Answer 37

if we have an equality, we can multiply both side in same number or term.

Answer 38

I multiply both side in (x-1).x^2 to make a quadratic equation.

Answer 39

before, I determine the domain.
in fraction a/b , b could not equal to zero.

Answer 40

then we solve our quadratic equation.

Answer 41

thank youu!!:))