1. ## Solving equations #1

Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

3) Factor completely. x - x^1/2 -6

here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

x^2+x-36 then from here i can factor it out... so im thinking this is the answer since
(x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....

4) ok last one( the one i cant seem to solve)

$\displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{ x-3}}=\displaystyle{\frac{x-2}{x-3}}$

so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

i put everything on one side. then made it so each fraction had the same denominator so i got.

-3+1-x-2/x-3 -> -4-x / x-3

from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

-(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

help would be much appreciated! thanks!

2. Originally Posted by Nismo
Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

(3) Factor completely. x - x^1/2 -6

here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

x^2+x-36 then from here i can factor it out... so im thinking this is the answer since
(x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....

4) ok last one( the one i cant seem to solve)
$\displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{ x-3}}=\displaystyle{\frac{x-2}{x-3}}$

so i tried this problem about6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

i put everything on one side. then made it so each fraction had the same denominator so i got.

-3+1-x-2/x-3 -> -4-x / x-3

from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

-(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

help would be much appreciated! thanks!
For 3: Hint
let $y=x^{\frac{1}{2}}$ this gives
$y^2-y-6$

3. Hey thanks for the help!

for problem #3 im still a bit confused. Although i think i was able to solve the problem and got the answer of 2x-1 im not comprehending on how the steps where done. i do know and realize you substitued x^1/2 for y and solved it that way. but im puzzled as to what happened to that regular x did that get subsituted for the y^2 or somthing?

4. Originally Posted by Nismo
4) ok last one( the one i cant seem to solve)

$\displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{ x-3}}=\displaystyle{\frac{x-2}{x-3}}$

so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

i put everything on one side. then made it so each fraction had the same denominator so i got.

-3+1-x-2/x-3 -> -4-x / x-3

from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

-(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

help would be much appreciated! thanks!
$\displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{ x-3}}=\displaystyle{\frac{x-2}{x-3}}$

Multiply that through by the LCD [which is 2(x-3)] - this is to clear the denominator and give a linear equation

$\left(\dfrac{1}{2} \cdot 2(x-3)\right) + \left(\dfrac{1}{x-3} \cdot 2(x-3)\right) = \left(\dfrac{x-2}{x-3} \cdot 2(x-3)\right)$

You can simplify to give $(x-3) + (2) = 2(x-2)$ which should be simple enough to solve

5. Originally Posted by Nismo
Hey thanks for the help!

for problem #3 im still a bit confused. Although i think i was able to solve the problem and got the answer of 2x-1 im not comprehending on how the steps where done. i do know and realize you substitued x^1/2 for y and solved it that way. but im puzzled as to what happened to that regular x did that get subsituted for the y^2 or somthing?

As TheEmptySet has suggested:

$y^2-y-6$ can be factorised as:

$y^2-3y+2y-6\;=\;......$

for your question no. 4. start by multiplying both sides of the equation by $x-3$

6. Originally Posted by Nismo
Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

3) Factor completely. x - x^1/2 -6

here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

x^2+x-36 then from here i can factor it out
"Rooted the whole problem"? If you mean by that $x-\sqrt{x}- 6= \sqrt{x^2+ x- 36}$,that is simply not true. The square root of $a^2- b^2$ NOT equal to $a- b$. For example, $5^2- 4^2= 25- 16= 9$ and $\sqrt{9}= 3$, not 5- 4= 1.

... so im thinking this is the answer since
(x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....
Instead, let $y= \sqrt{x}$ so that $x= y^2$. Now you have $y^2- y- 6= (y- 3)(y+ 2)= 0$. Setting $y= \sqrt{x}= x^{1/2}$ again, that tells us that $x- x^{1/2}- 6= (x^{1/2}- 3)(x^{1/2}+ 2)$.

Once again, $\sqrt{a+ b}$ is NOT $\sqrt{a}+ \sqrt{b}$. That was your mistake.

4) ok last one( the one i cant seem to solve)

$\displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{ x-3}}=\displaystyle{\frac{x-2}{x-3}}$

so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

i put everything on one side. then made it so each fraction had the same denominator so i got.

-3+1-x-2/x-3 -> -4-x / x-3

from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

-(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

help would be much appreciated! thanks!