Algebra II Equations Containing Radicals Part 2

[velox_xox]

I managed to get the other problems right, but this one I've been fiddling with and can't seem to get the right answer.

Homework Statement

$$\sqrt {3b -2} - \sqrt {2b + 5} = 1$$

Answer: $22$

Homework Equations

--

The Attempt at a Solution

For this I tried the method of leaving the equation as is and squaring both sides, since this is the way the textbook seems to want me to learn it for this lesson.

$\sqrt {3b - 2} - \sqrt {2b + 5} = 1$
$(\sqrt {3b - 2} - \sqrt {2b + 5})^2 = 1^2$
$3b - 2 - 2\sqrt {6b^2 - 10} + 2b + 5 = 1$
$5b + 3 - 2\sqrt {6b^2 - 10} = 1$
$5b + 3 = 2\sqrt {6b^2 - 10}$
$(5b + 3)^2 = (2\sqrt {6b^2 - 10})^2$
$25b^2 + 30b + 9 = 4(6b^2 - 10)$
$25b^2 + 30b + 9 = 24b^2 - 40$
$b^2 + 30b +49 = 0$

At this point, I realize that 22 isn't a factor of that final equation. I've also tried the other methods, such as $\sqrt {3b - 2} = \sqrt {2b + 5}$, and I get b = 7.

I've tried to look through it for obvious arithmetic mistakes or my tendency to create incorrect products of the form (a + b)^2, but I didn't see anything. It could be the sheer volume of the problem has me spooked or tripping up.

Any help would be greatly appreciated. Thanks in advance.

 

[Mark44]

I managed to get the other problems right, but this one I've been fiddling with and can't seem to get the right answer.

Homework Statement

$$\sqrt {3b -2} - \sqrt {2b + 5} = 1$$

Answer: $22$

Homework Equations

--

The Attempt at a Solution

For this I tried the method of leaving the equation as is and squaring both sides, since this is the way the textbook seems to want me to learn it for this lesson.

$\sqrt {3b - 2} - \sqrt {2b + 5} = 1$
$(\sqrt {3b - 2} - \sqrt {2b + 5})^2 = 1^2$
$3b - 2 - 2\sqrt {6b^2 - 10} + 2b + 5 = 1$
$5b + 3 - 2\sqrt {6b^2 - 10} = 1$
$5b + 3 = 2\sqrt {6b^2 - 10}$
$(5b + 3)^2 = (2\sqrt {6b^2 - 10})^2$
$25b^2 + 30b + 9 = 4(6b^2 - 10)$
$25b^2 + 30b + 9 = 24b^2 - 40$
$b^2 + 30b +49 = 0$

At this point, I realize that 22 isn't a factor of that final equation. I've also tried the other methods, such as $\sqrt {3b - 2} = \sqrt {2b + 5}$, and I get b = 7.

This is how I would do it, but you have omitted a term.

The equation should be $\sqrt {3b - 2} = \sqrt {2b + 5} + 1$.
Now square both sides.

I've tried to look through it for obvious arithmetic mistakes or my tendency to create incorrect products of the form (a + b)^2, but I didn't see anything. It could be the sheer volume of the problem has me spooked or tripping up.

Any help would be greatly appreciated. Thanks in advance.

 

[Hertz]

Starting from the original equation, move the second radical to the other side and square both sides. You should end up with:

$3b - 2 = (1 + \sqrt{2b + 5})^2$

After foiling, combine like terms and solve for the $2\sqrt{2b + 5}$ term. You should end up with b - 8 on the other side.

Square both sides once more.

$(b - 8)^2 = 4(2b + 5)$

Foil/distribute and set equal to zero.

Factor the remaining equation and you should end up with zeroes of 2 and 22.
Plug both answers back into the original equation to see which ones work.

Good luck :)

 

[Bohrok]

$\sqrt {3b - 2} - \sqrt {2b + 5} = 1$
$(\sqrt {3b - 2} - \sqrt {2b + 5})^2 = 1^2$
$3b - 2 - 2\sqrt {6b^2 - 10} + 2b + 5 = 1$

You're missing the +11b term from multiplying out (3b-2)(2b+5).

 

[Villyer]

$5b + 3 - 2\sqrt {6b^2 - 10} = 1$
$5b + 3 = 2\sqrt {6b^2 - 10}$

You also forgot to subtract the 1 on the right side over to the left side.

 

[velox_xox]

Aaah, I see. Apparently, I have trouble with my factoring instincts. A.k.a. I need to factor things out the old fashioned way as opposed to doing mental math.

I see where I went wrong and even though my textbook seems to want me to keep the radicals all on one side, I think I'm going to make exceptions for cases such as this. It gets incredibly messy and too advanced for me the other way. Nevertheless, I tried it out just to try and understand what I was missing; and even though, I couldn't arrive at an answer that way, thanks to everyone's pointers I see where I went wrong here again.

So, thank you Mark44, Hertz (thanks for the luck; I need it! :D), Bohrok, and Villyer.
Here's what I got the last time:

$\sqrt {3b -2} - \sqrt {2b + 5} = 1$
$(\sqrt {3b - 2})^2 =(\sqrt {2b + 5}+ 1)^2$
$3b - 2 = 2b + 5 + 2\sqrt {2b + 5} + 1$
$(b - 8)^2 = (2\sqrt {2b + 5})^2$
$b^2 - 16b + 64 = 4(2b + 5)$
$b^2 -16b + 64 = 8b + 20$
$b^2 - 24b + 44 = 0$
$(b - 2)(b - 22) = 0$
b = 2 or b = 22

'2' doesn't work; it produces 2 - 3 = 1; '22' checks out with an 8 - 7 = 1. So, unless I bent the rules of algebra to solve this, case closed.

Thanks once again everyone!