# Solving equations containing square roots.

• Jan 31st 2010, 07:31 PM
Anemori
Solving equations containing square roots.
a) $\sqrt{2x+1} - \sqrt{x} = 1$
True or false?

b) $\sqrt{x+2} + \sqrt{7x+2} = 6$
True or false?

• Jan 31st 2010, 07:47 PM
halpo
Quote:

Originally Posted by Anemori
a) $\sqrt{2x+1} - \sqrt{x} = 1$
True or false?

b) $\sqrt{x+2} + \sqrt{7x+2} = 6$
True or false?

a) $\sqrt{2x+1} - \sqrt{x} = 1$
True or false?

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

Square both sides...

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

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

$x=2\sqrt{x}$

We see that $x=0$

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

So this is a true statement.

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

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

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

Square both sides...

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

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

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

Look, we can divide the equation by -6.

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

Square both sides again...

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

Distribute the 4.

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

Subtract 28x and 8 from the right.

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

This factors to...

$(x-14)(x-2)$

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 PM
Anemori
Thank you that was awsome....
• Feb 5th 2010, 12:30 AM
Anemori
Quote:

Originally Posted by halpo
a) $\sqrt{2x+1} - \sqrt{x} = 1$
True or false?

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

Square both sides...

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

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

$x=2\sqrt{x}$

We see that $x=0$

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

So this is a true statement.

You can try B...

i was confused with this one.. how come x = 0 on this $x=2\sqrt{x}$ im sorry dumb question.. i just dont get it..
• Feb 5th 2010, 01:47 AM
HallsofIvy
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 $\sqrt{x}$, you square to get rid of the square roots. After getting to $x= 2\sqrt{x}$, square again to get $x^2= 4x$ which is the same as $x^2- 4x= x(x- 4)= 0$. 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 $\sqrt{2(0)+1}- \sqrt{0}= \sqrt{1}= 1$ so x= 0 is a solution. With x= 4, it becomes $\sqrt{2(4)+ 1}- \sqrt{4}= \sqrt{9}- \sqrt{4}= 3- 2= 1$ so x= 4 is also a solution.