I tried to solve those euqations but made errors. Could you please solve those irrational euqations!
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√4x+9 - 2√x = 1
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√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!
<-- Square both sides:
<-- Square both sides again:
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
Okay;
The first was already answered;
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.
Final answer: x = 4
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 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...
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.
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?
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):
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