Help with equations

Ximei

New member
Joined
Jan 26, 2010
Messages
5
I hope there is someone out there who can help me. :?
I am trying to solve this equation:
x+5+x+3=2x+7\displaystyle \sqrt{x+5}+\sqrt{x+3}=\sqrt{2x+7}
I already found out that x=>-3
I tryed to solve it and got here:
2(x+5)(x+3)=1\displaystyle 2\sqrt{(x+5)(x+3)}= -1
I don't know how to solve it from here ,every method I try gives me some weird number.I’d appreciate an answer along with the steps to solve it.
And I also have to solve this one, but I don't even know from where to start :(
2x25x+2x2x2=x23x+2\displaystyle \sqrt{2x^2 -5x+2}- \sqrt{ x^2-x-2}= \sqrt{x^2-3x+2}
Maybe someone can give me a hint.
 
Hello, Ximei!

The second one is strange . . . never had one like it!


2x25x+2x2x2=x23x+2\displaystyle \sqrt{2x^2 -5x+2}- \sqrt{ x^2-x-2}\:=\: \sqrt{x^2-3x+2}

Those polynomials just happen to factor . . .


We have:   (2x1)(x2)(x+1)(x2)  =  (x1)(x2)\displaystyle \text{We have: }\;\sqrt{(2x-1)(x-2)} - \sqrt{(x+1)(x-2)} \;=\;\sqrt{(x-1)(x-2)}

. . . . . . . . .2x1 ⁣ ⁣x2x+1 ⁣ ⁣x2  =  x1 ⁣ ⁣x2\displaystyle \sqrt{2x-1}\!\cdot\!\sqrt{x-2} - \sqrt{x+1}\!\cdot\!\sqrt{x-2} \;=\;\sqrt{x-1}\!\cdot\!\sqrt{x-2}

And we see that x=2 is a solution.\displaystyle \text{And we see that }\,\boxed{x = 2}\,\text{ is a solution.}



If x2, we can divide by x2 ⁣:2x1x+1  =  x1\displaystyle \text{If }x\neq2\text{, we can divide by }\sqrt{x-2}\!:\quad \sqrt{2x-1} - \sqrt{x+1} \;=\;\sqrt{x-1}

. . And this equation can be solved in the "usual way".

 
Top