how do i solve sqrt(x+1)+sqrt(2x)=sqrt(5x+3)

Question

mathstudent55 · Answer

You need to square both sides.
Then combine like terms.
Move the root to one side, and square both sides again.
Finally, check all solutions in the original equation bec
squaring both sides may introduce extraneous solutions.

mathstudent55 · Answer

The above solution is incorrect.
The square of the left side is the square of a binomial.
$(a + b)^2 \ne a^2 + b^2$

mathstudent55 · Answer

$ (\sqrt{x+1}+\sqrt{2x})^2=(\sqrt{5x+3})^2 $

The left side has to be squared as the square of a binomial. There will be a middle term containing a square root. That is why you will need to square both sides again.

anonymous · Answer

(sqrt(x+1)+2x)(sqrt(x+2(2x))=5x+3
2(sqrt(2x^2+2x)=2x+2
sqer2z^2+2x=x+1
2x^2+2x=x^2+2x+1
x^2-1=0
x^-1=0
(x+1)(x-1)=0
x=-1,1

anonymous · Answer

@johnryannn  mathstudent is right. This isn't a good sign for my log test tommorow

mathstudent55 · Answer

$(\sqrt{x+1}+\sqrt{2x})^2=(\sqrt{5x+3})^2 $

$  (\sqrt{x+1})^2 + 2 (\sqrt{x+1})(\sqrt{2x}) +( \sqrt{2x})^2 = 5x + 3$

$x + 1 + 2\sqrt{(x + 1)(2x)} + 2x = 5x + 3$

$3x + 1 + 2\sqrt{2x^2 + 2x}  = 5x + 3$

$2\sqrt{2x^2 + 2x} = 2x + 2$

$(2\sqrt{2x^2 + 2x})^2 = (2x + 2)^2$

$4(2x^2 + 2x) = 4x^2 + 8x + 4$

$2(2x^2 + 2x) = 2x^2 + 4x + 2$

$4x^2 + 4x = 2x^2 + 4x + 2$

$2x^2 - 2 = 0$

$x^2 - 1 = 0$

$x + 1 = 0 $   or   $x - 1 = 0$

$x = -1 $   or  $x = 1$

Now x = -1 and x = 1 must be checked in the original equation.

Check x = -1:

$\sqrt{x+1}+\sqrt{2x}=\sqrt{5x+3} $

$\sqrt{-1+1}+\sqrt{2(-1)}=\sqrt{5(-1)+3} $

$ \sqrt{0}+\sqrt{-2}=\sqrt{-2} $

Since this involves the square root of negative numbers, this solution is discarded.

Check x = 1:

$\sqrt{x+1}+\sqrt{2x}=\sqrt{5x+3} $

$\sqrt{1+1}+\sqrt{2(1)}=\sqrt{5(1)+3} $

$\sqrt{2}+\sqrt{2}=\sqrt{8} $

$2\sqrt{2}=2\sqrt{2} $

The solution x = 1 is valid.

The final solution is: x = 1.

anonymous · Answer

thanks

mathstudent55 · Answer

You're welcome.