# Irrational euqations!!!

• Jan 15th 2007, 01:01 PM
map
Irrational euqations!!!
I tried to solve those euqations but made errors. Could you please solve those irrational euqations!

____ __
√4x+9 - 2√x = 1
____ ____
√3x+4+ √x-3 = 5

I tried to write that 4x+9 all together is under √ sign not only 4x and same for 3x+4. I hope you understand what I mean!
• Jan 15th 2007, 01:06 PM
AfterShock
Quote:

Originally Posted by map
I tried to solve those euqations but made errors. Could you please solve those irrational euqations!

_____ __
√4x+9 - 2√x = 1
____ ___
√3x+4+ √x-3 = 5

I have no clue what that is. I will assume it is as follows:

sqrt(4x + 9) - 2*sqrt(x) = 1 and

sqrt(3x + 4) + sqrt(x-3) = 5 ?

Or is it all under the sqrt? Write it the way I have written it, or use LaTeX.
• Jan 15th 2007, 01:12 PM
map
4x + 9 are under sqrt, not only 4x

3x + 4 are under sqrt, not only 3x
• Jan 15th 2007, 01:29 PM
topsquark
Quote:

Originally Posted by map
____ __
√4x+9 - 2√x = 1

$\displaystyle \sqrt{4x + 9} - 2\sqrt{x} = 1$

$\displaystyle \sqrt{4x + 9} = 2\sqrt{x} + 1$ <-- Square both sides:

$\displaystyle 4x + 9 = (2\sqrt{x} + 1)^2$

$\displaystyle 4x + 9 = 4x + 4\sqrt{x} + 1$

$\displaystyle 8 = 4\sqrt{x}$

$\displaystyle \sqrt{x} = 2$ <-- Square both sides again:

$\displaystyle x = 4$

Now, whenever we square both sides of the equation we should ALWAYS check the solution in the original equation. (Squaring both sides of an equation can add extra solutions.) As it turns out, x = 4 does work.

-Dan
• Jan 15th 2007, 01:44 PM
topsquark
Quote:

Originally Posted by map
____ ____
√3x+4+ √x-3 = 5

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

$\displaystyle \sqrt{3x + 4} = 5 - \sqrt{x - 3}$ <-- Square both sides:

$\displaystyle 3x + 4 = (5 - \sqrt{x - 3} )^2$

$\displaystyle 3x + 4 = 25 - 10\sqrt{x - 3} + (x - 3)$

$\displaystyle 2x - 18 = -10\sqrt{x - 3}$ <-- Square both sides again:

$\displaystyle (2x - 18)^2 = 100(x - 3)$

$\displaystyle 4x^2 - 72x + 324 = 100x - 300$

$\displaystyle 4x^2 - 172x + 624 = 0$

$\displaystyle 4(x^2 - 43x + 156) = 0$

$\displaystyle 4(x - 4)(x - 39) = 0$

Thus x = 4 and x = 39.

Again, we must check these solutions in the original equation. It turns out that only x = 4 will work.

Note the process that was followed in both of these: isolate one of the square roots on one side of the equation, then square both sides. Then isolate the remaining square root and square both sides again. If you have more than two radical expressions you can repeat this process as many times as necessary.

-Dan
• Jan 15th 2007, 01:47 PM
AfterShock
Quote:

Originally Posted by map
4x + 9 are under sqrt, not only 4x

3x + 4 are under sqrt, not only 3x

Okay;

The second:

sqrt(3x + 4) + sqrt(x - 3) = 5

5 - sqrt(x - 3) = sqrt(3x + 4)

Square each side;

(5 - sqrt(x - 3))^2 = 3x + 4

(5 - sqrt(x - 3))*(5 - sqrt(x - 3)) = 3x + 4

25 - 5sqrt(x - 3) - 5sqrt(x - 3) + sqrt(x - 3)^2 = 3x + 4

25 - 10sqrt(x - 3) + x - 3 = 3x + 4

-10sqrt(x - 3) = 2x - 18

Square both sides;

100*(x - 3) = (2x - 18)^2

100x - 300 = 4x^2 -72x +324

4x^2 -172x + 624 = 0

4(x - 39)*(x - 4) = 0

x = 39 or x = 4

x = 39 does not work, therefore throw that out; check to see that x = 4 works.

It does.

• Jan 15th 2007, 02:09 PM
map
Thanks for answers but I don't wan't to bother you but I rechecked answers and in second example I couldn't get asnwer 4 only 39 and 2... I used this formula D=b*b-4*a*c and then x=-bą√D/2*a

Where is my mistake?
• Jan 15th 2007, 02:11 PM
map
sorry my mistake you're right! :cool:
• Jan 15th 2007, 03:47 PM
topsquark
Quote:

Originally Posted by map
Thanks for answers but I don't wan't to bother you but I rechecked answers and in second example I couldn't get asnwer 4 only 39 and 2... I used this formula D=b*b-4*a*c and then x=-bą√D/2*a

Where is my mistake?

Quote:

Originally Posted by map
sorry my mistake you're right! :cool:

Either way, thanks for checking. Someone has to keep us honest! :D

-Dan
• Jan 21st 2007, 12:44 PM
map
OK still trying to improve my skills everything is going quiet nice although not too easy any way I tried to do few other easy looking exercises and... errors :mad: OK I won't give up anyway but could you please help me on few exercises?
1st one:

sqrt(3x + 4) + sqrt(x + 2) = 8

Everything was OK but then I got error went back to recheck but didn't find anything, then rechecked next day still nothing... :mad: :mad:

P.s. How it's possible write it like topsquark is it that LaTeX?
• Jan 21st 2007, 03:49 PM
AfterShock
Quote:

Originally Posted by map
OK still trying to improve my skills everything is going quiet nice although not too easy any way I tried to do few other easy looking exercises and... errors :mad: OK I won't give up anyway but could you please help me on few exercises?
1st one:

sqrt(3x + 4) + sqrt(x + 2) = 8

Everything was OK but then I got error went back to recheck but didn't find anything, then rechecked next day still nothing... :mad: :mad:

P.s. How it's possible write it like topsquark is it that LaTeX?

sqrt(3x + 4) + sqrt(x + 2) = 8

sqrt(x + 2) = 8 - sqrt(3x + 4)

Square both sides;

x + 2 = (8 - sqrt(3x + 4))^2

x + 2 = (8 - sqrt(3x + 4))*(8 - sqrt(3x + 4))

x + 2 = 68 - 8sqrt(3x + 4) - 8sqrt(3x + 4) + 3x

x + 2 = 68 - 2*8*sqrt(3x + 4) + 3x

x + 2 = 68 - 16sqrt(3x + 4) + 3x

x - 66 = -16sqrt(3x + 4) + 3x

(-2x - 66)/(-16) = sqrt(3x + 4)

Square both sides;

(-2x - 66)^2/256 = 3x + 4

(-2x - 66)^2 = 768x + 1024

4x^2 + 264x + 4356 = 768x + 1024

4x^2 - 504x + 3332 = 0

4(x - 119)*(x - 7) = 0

x = 7 V x = 119

Go back to the original and check:

sqrt(3x + 4) + sqrt(x + 2) = 8

sqrt(3(7) + 4) + sqrt(7 + 2) = 8 ? Yes.

sqrt(3(119) + 4) + sqrt(119 + 2) ? No.

Thus, the final answer is x = 7.
• Jan 22nd 2007, 10:01 AM
map
Hello I tried to solve those euqations and I tried to found out more easy way how to resolve them faster and without big figures. so one guy showed this methode he us. He said that many teachers won't do it in this way but it works. So I tried ...and it works .... What do you think? Is this correct way how to solve those euqations.

sqrt(x + 2) + sqrt(x - 2)=sqrt(3x - 2)

sqrt(x + 2) = sqrt(3x - 2) - sqrt(x - 2)

sqrt(x + 2) = sqrt(3x - 2 - x + 2)

sqrt(x + 2) = sqrt(2x) then square both sides.

x + 2 = 2x

-x = -2

x = 2

Check:

sqrt(2 + 2) + sqrt(2 - 2) ? sqrt(6 - 2)

sqrt(2 + 2 + 2 - 2) ? sqrt(4)

sqrt(4) = sqrt(4)

2=2

I did it also in usual way and I got same answer,so...? What do you think?
• Jan 22nd 2007, 01:06 PM
topsquark
Quote:

Originally Posted by map
Hello I tried to solve those euqations and I tried to found out more easy way how to resolve them faster and without big figures. so one guy showed this methode he us. He said that many teachers won't do it in this way but it works. So I tried ...and it works .... What do you think? Is this correct way how to solve those euqations.

sqrt(x + 2) + sqrt(x - 2)=sqrt(3x - 2)

sqrt(x + 2) = sqrt(3x - 2) - sqrt(x - 2)

sqrt(x + 2) = sqrt(3x - 2 - x + 2)

OUCH!!! I think this is either a trick problem or that your friend needs to pay more attention to the details. Look at the RHS for the last two lines. For arbitrary a and b (such that a, b, and a - b are positive):
$\displaystyle \sqrt{a} - \sqrt{b} \neq \sqrt{a - b}$

Just pick a couple of a and b values to prove it to yourself. So either the problem was designed to give a correct result despite the error in the Mathematics, or your friend was very lucky indeed to get the correct answer using this method.

-Dan
• Jan 22nd 2007, 01:20 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
OUCH!!! I think this is either a trick problem or that your friend needs to pay more attention to the details. Look at the RHS for the last two lines. For arbitrary a and b (such that a, b, and a - b are positive):
$\displaystyle \sqrt{a} - \sqrt{b} \neq \sqrt{a - b}$

Just pick a couple of a and b values to prove it to yourself. So either the problem was designed to give a correct result despite the error in the Mathematics, or your friend was very lucky indeed to get the correct answer using this method.

-Dan

I find it quite funny many students think that,
$\displaystyle (a+b)^n=a^n+b^n$.
$\displaystyle (a+b)^p=a^p+b^p$.