# Math Help - radical equation

Dnt know how to work these out:
$\sqrt{\frac{2}{5}x + 3} - \frac{4}{5} = \frac{2}{5}$

$6 + \sqrt{7x+3} = 6$

2. $
\sqrt{\frac{2}{5}x + 3} - \frac{4}{5} = \frac{2}{5}
$

Add $\frac{4}{5}$ to both sides of the equation:

$
\sqrt{\frac{2}{5}x + 3} = \frac{6}{5}
$

Square both sides of the equation:

$\frac{2}{5}x + 3 = \frac{36}{25}$

Subtract 3 from both sides:

$\frac{2}{5}x = -\frac{39}{25}$

Multiply both sides by 5:

$2x = -\frac{195}{25}$

Divide both sides by 2:

$x = -\frac{195}{50} = -\frac{39}{10}$

3. $
6 + \sqrt{7x+3} = 6
$

Subtract 6 from both sides:

$\sqrt{7x+3} = 0$

Square both sides:

$7x+3 = 0$

Subtract 3 from both sides:

$7x = -3$

Divide both sides by 7:

$x = -\frac {3}{7}$

4. So on a problem like this: $4- \sqrt{2x+3} = 5$ do I add 4 to both sides?

And on this one, I add 2 to both sides then subtract 5? Not sure since there isn't an addition sign after it like the problem before. $5 \sqrt{\frac{4x}{3}} - 2 = 0$

5. So on a problem like this: $4- \sqrt{2x+3} = 5$ do I add 4 to both sides?
No. If you added 4 to both sides you would not simplify anything at all. Try rearranging the equation first like this:

$- \sqrt{2x+3} + 4 = 5$

Now do you see that you have to subtract 4 from both sides?

6. Originally Posted by Belanova
So on a problem like this: $4- \sqrt{2x+3} = 5$ do I add 4 to both sides? Mr F says: Noooo. You >subtract< 4 from both sides.

And on this do I subtract? Not sure since there isn't an addition sign after it like the problem before. $5 \sqrt{\frac{4x}{3}} - 2 = 0$ Mr F says: You >add< 2 to both sides.
..

7. And on this one, I add 2 to both sides then subtract 5? Not sure since there isn't an addition sign after it like the problem before. $5 \sqrt{\frac{4x}{3}} - 2 = 0$
You would add 2 to both sides, but instead of subtracting 5 you would divide both sides by 5.

8. You try to isolate the square root expression all by itself as seen in your previous two questions. That way, you can simply square both sides without having to go through too much algebra.

For the first one:
$4 - \sqrt{2x + 3} = 5$
$\sqrt{2x + 3} = -1$

You can square both sides much more easily than if you did it to the first line.

For the second one, again isolate the square root expression to one side so you can square it.

9. $4- \sqrt{2x+3} = 5$
$= -1 = no solution$

$5 \sqrt{\frac{4x}{3}} - 2 = 0$
$= \frac{3}{25}$

10. $4- \sqrt{2x+3} = 5
= -1 = no solution$
No not quite. To get rid of a square root, you do not take the square root of it, you square it. So, instead of taking the square root of -1, square -1.

Then you have that:

$2x+3 = 1$

Subtract 3 from both sides:

$2x = -2$

Divide both sides by 2:

$x = -1$

11. $5 \sqrt{\frac{4x}{3}} - 2 = 0
= \frac{3}{25}$
Yes, that is correct.

12. Originally Posted by topher0805
No not quite. To get rid of a square root, you do not take the square root of it, you square it. So, instead of taking the square root of -1, square -1.

Then you have that:

$2x+3 = 1$

Subtract 3 from both sides:

$2x = -2$

Divide both sides by 2:

$x = -1$

You made a mortal mistake there. After each step, check the domain and the range! Especially after taking the square of both sides!

You say that,
$4 -\sqrt{2x+3} = 5$

$\sqrt{2x+3} = -1$

We could just take the square of both sides and go on and find x=-1.. If $\sqrt{2x+3}$ hadn't had to be equal to or greater than zero..

You probably made this mistake because you didn't notice that $\sqrt{x}$ is not the inverse function of $x^2$.

$\sqrt{x}$ is never negative.