a) $\displaystyle \sqrt{2x+1} - \sqrt{x} = 1 $

True or false?

b) $\displaystyle \sqrt{x+2} + \sqrt{7x+2} = 6 $

True or false?

Can you please help me understand this.

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- Jan 31st 2010, 07:31 PMAnemoriSolving equations containing square roots.
a) $\displaystyle \sqrt{2x+1} - \sqrt{x} = 1 $

True or false?

b) $\displaystyle \sqrt{x+2} + \sqrt{7x+2} = 6 $

True or false?

Can you please help me understand this. - Jan 31st 2010, 07:47 PMhalpo
a) $\displaystyle \sqrt{2x+1} - \sqrt{x} = 1 $

True or false?

$\displaystyle \sqrt{2x+1}=(1+\sqrt{x})$

Square both sides...

$\displaystyle 2x+1=1+2\sqrt{x}+x$

Subtract $\displaystyle 1$ and $\displaystyle x$ from the right...

$\displaystyle x=2\sqrt{x}$

We see that $\displaystyle x=0$

Plugging it in shows that $\displaystyle \sqrt{0+1}-\sqrt{0}=1$

So this is a true statement.

You can try B... - Jan 31st 2010, 08:07 PMhalpo
Here's B anyway since it's more involved....

$\displaystyle \sqrt{x+2} + \sqrt{7x+2} = 6 $

$\displaystyle (\sqrt{x+2})=(6-\sqrt{7x+2})$

Square both sides...

$\displaystyle x+2=36-12\sqrt{7x+2}+7x+2$

Subtract 36, 7x, and 2 from the right.

$\displaystyle -6x-36=-12\sqrt{7x+2}$

Look, we can divide the equation by -6.

$\displaystyle (x+6)=(2\sqrt{7x+2})$

Square both sides again...

$\displaystyle x^2+12x+36=4(7x+2)$

Distribute the 4.

$\displaystyle x^2+12x+36=28x+8$

Subtract 28x and 8 from the right.

$\displaystyle x^2-16x+28=0$

This factors to...

$\displaystyle (x-14)(x-2)$

So our solutions are 14 and 2. Check them! You will find that 2 is the only solution and thus the statement is true. - Jan 31st 2010, 08:27 PMAnemori
Thank you that was awsome....

- Feb 5th 2010, 12:30 AMAnemori
- Feb 5th 2010, 01:47 AMHallsofIvy
Were those really "true or false" questions? I would interpret "true or false" for an equation involving, say, x, to be asking if it was true for

**all**x- and that's clearly false for both questions.

To solve equations involving $\displaystyle \sqrt{x}$, you**square**to get rid of the square roots. After getting to $\displaystyle x= 2\sqrt{x}$, square again to get $\displaystyle x^2= 4x$ which is the same as $\displaystyle x^2- 4x= x(x- 4)= 0$. Either x= 0 or x- 4= 0 so x= 4.

Since squaring both sides like that can introduce "extraneous roots", you should always check by putting the numbers into the original equation. The original equation was [tex]\sqrt{2x+1}- \sqrt{x}= 1[tex]. With x= 0 that becomes $\displaystyle \sqrt{2(0)+1}- \sqrt{0}= \sqrt{1}= 1$ so x= 0 is a solution. With x= 4, it becomes $\displaystyle \sqrt{2(4)+ 1}- \sqrt{4}= \sqrt{9}- \sqrt{4}= 3- 2= 1$ so x= 4 is also a solution.