I can't figure out this radical equation?

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October 13th 2009, 12:43 PM
bwenner

I can't figure out this radical equation?

I'm taking Pre-Calc and Trigonometry, and the class is so poorly put together I don't even get a textbook. My teacher is of no help. I can't figure out how to solve this problem, and I desperately would like to understand. Help?

√x+2 + √3x+4 = 2

I have tried it over and over and have gotten different answers each time. I don't know what are the proper steps to solve an equation with two radicals on one side. :<
October 13th 2009, 12:57 PM
pickslides

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

Grouping like terms together

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

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

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

Factoring

$\sqrt{x} (1+\sqrt{3}) = -4$

$\sqrt{x} = \frac{-4}{ (1+\sqrt{3}) }$

Now square both sides
October 13th 2009, 01:00 PM
bwenner

Quote:

Originally Posted by pickslides

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

Grouping like terms together

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

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

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

Factoring

$\sqrt{x} (1+\sqrt{3}) = -4$

$\sqrt{x} = \frac{-4}{ (1+\sqrt{3}) }$

Now square both sides

I get that, but the +2 and +4 are both supposed to be under the square root symbols. I wasn't sure how to express that. Maybe √(x+2) + √(3x+4) = 2 ? Sorry for the confusion, and thanks so much for the help!
October 13th 2009, 02:14 PM
Krizalid

Quote:

Originally Posted by pickslides

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

you may notice that having this the obvious conclusion is that there's no solution here, so it leads to think that the equation was stated as bwenner says.
October 13th 2009, 03:11 PM
pickslides

$\sqrt{1+x}= 1+\frac{1}{2}x- \frac{1}{8}x^2+\frac{1}{16}x^3-\dots$

Might be outside of the scope of what you are learning but you can try this series approximation?

Using it as a 2nd order approximation could be easy to solve.
October 13th 2009, 04:29 PM
skeeter

Quote:

Originally Posted by bwenner

√(x+2) + √(3x+4) = 2

note the use of parentheses to make your equation clear.

start by squaring both sides ...

$(x+2) + 2\sqrt{x+2}\sqrt{3x+4} + (3x+4) = 4$

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

$\sqrt{x+2}\sqrt{3x+4} = -(2x+1)$

square again ...

$(x+2)(3x+4) = 4x^2 + 4x + 1$

$3x^2 + 10x + 8 = 4x^2 + 4x + 1$

$0 = x^2 - 6x -7$

$0 = (x - 7)(x + 1)$

$x = 7$ or $x = -1$

be sure to check the validity of both values in the original equation to rule out extraneous solutions.
October 13th 2009, 04:51 PM
pacman

Or this might be the right form? √(x+2) + √(3x+4) = 2.

squaring BS,

(x + 2) + 2√(x + 2)√(3x + 4) + 3x + 4 = 4, simplify

2√(x + 2)√(3x + 4) = 4 - 4 - 2 - x - 3x,

2√(x + 2)√(3x + 4) = -2 -4x = -2(1 + 2x)

√(x + 2)√(3x + 4) = -(1 + 2x), square BS,

(x + 2)(3x + 4) = (-1 - 2x)^2,

3x^2 + 10 x + 8 = 1 + 4x + 4x^2, combine like terms

(4x^2 - 3x^2) + (-10x + 4x) + (1 - 8) = 0,

x^2 - 6x - 7 = 0, factoring

(x - 7)(x + 1) = 0, using the zero factor theorem

x = -1 or x = 7.

Looking for extraneous root, we substitute the value to the original equation.

(x + 2)^(1/2) + (3x + 4)^(1/2) = 2, for x = 7

(7 + 2)^(1/2) + (3(7) + 4)^(1/2) = 2?

3 + 5 = 2? not the root of the original equation,

Next, try x = -1.

(- + 2)^(1/2) + (3(-1) + 4)^(1/2) = 2?

1 + 1 = 2. x = -1 is the ONLY ROOT of the above equation.

Check by graphing: surely x = -1 is a ROOT.

(Bow)
October 13th 2009, 04:54 PM
pacman

1 Attachment(s)

see the graph, x = -1 is the only solution.
October 13th 2009, 05:00 PM
Defunkt

Just a note, it's easy enough to see that the function may only have one real root, seeing as it is monotonous increasing.
October 14th 2009, 04:34 AM
HallsofIvy

I would separate the two square roots:
$\sqrt{x+2} + \sqrt{3x+4} = 2$
$\sqrt{x+2}= 2- \sqrt{3x+4}$

Now square both sides
$x+ 2= 2- 2\sqrt{3x+4}+ 2x+ 4$
$2\sqrt{3x+4}= x+ 4$

Squaring again,
$2(3x+4)= (x+4)^2= x^2+ 8x+ 4$
$x^2+ 8x+ 4= 6x+ 8$
$x^2+ 2x- 4= 0$

Can you finish from there? Remember to check any roots in the origina equation. Squaring both sides can introduce "extraneous" roots.
October 14th 2009, 07:34 AM
pacman

HallsofIvy: "http://www.mathhelpforum.com/math-he...7f5e9446-1.gif. - here
Now square both sides http://www.mathhelpforum.com/math-he...d6fe2c81-1.gif
http://www.mathhelpforum.com/math-he...094968ff-1.gif""

you missed to square 2.