i need help?

Question

square root of (x-9) + square root of (x) = 1

and

2 (square root of (x-3)) + square root of (3x-5) = 8

Answer 1

Hey, a fellow 'need help'er!! :)

square both sides...
x-9+2sqrt(x^2-9x)+x=1
2x-10=2sqrt(x^2-9x)
x-5=sqrt(x^2-9x)

square both sides again
x^2-10x+25=x^2-9x
x=25

Do the exact same thing for the second equation. You will get x = 7

Answer 2

Presuming that what you're supposed to do is solve for x AND that complex numbers are not allowed, there is no solution for the first equation: if x is less than 9, you get a complex number. If x is 9 or more, the value on the left side is 3 or greater, so there is no solution.

For the second, x=7 is the only solution.

Answer 3

Question #1:
square root of (x-9) + square root of (x) = 1
YOU NEED TO SQUARE ALL FIELDS ON BOTH SIDES, TO PRODUCE THIS:

(x-9) + x = 1
2x - 9 = 1
2x = 10
x = 5

If you do the same operation to Question 2, i.e. squaring all fields on both sides, you'll get this:
4(x-3) + (3x-5) = 64
4x - 12 + 3x - 5 = 64
7x -17 = 64
7x = 81
x = 81/7

Answer 4

First one:

Get one sq. root by itself:

sqrt(x-9) = 1 - sqrtx

Square both sides:

x-9 = (1 - sqrt(x))(1 - sqrt(x))

x-9 = 1 - 2sqrt(x) + x

Subtract x and 1 from both sides:

-10 = -2sqrt(x)

5= sqrt x

Square both sides:
25 = x

BUT: when you check this answer, it doesn't work, so there is NO SOLUTION. Always check your answer when you square both sides of an equation, because sometimes you will get extraneous roots.

Answer 5

Put the question in your subject line. I need help isn't a question.

Answer 6

So what's your question?

Answer 7

i need help too

Answer 8

so????????? whatz the question?