# I can't figure out this radical equation?

• Oct 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. :<
• Oct 13th 2009, 12:57 PM
pickslides
$\displaystyle \sqrt{x} +2 +\sqrt{3x}+4 = 2$

Grouping like terms together

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

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

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

Factoring

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

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

Now square both sides
• Oct 13th 2009, 01:00 PM
bwenner
Quote:

Originally Posted by pickslides
$\displaystyle \sqrt{x} +2 +\sqrt{3x}+4 = 2$

Grouping like terms together

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

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

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

Factoring

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

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

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!
• Oct 13th 2009, 02:14 PM
Krizalid
Quote:

Originally Posted by pickslides
$\displaystyle \sqrt{x} +\sqrt{3x} = -4$

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.
• Oct 13th 2009, 03:11 PM
pickslides
$\displaystyle \sqrt{1+x}= 1+\frac{1}{2}x- \frac{1}{8}x^2+\frac{1}{16}x^3-\dots$

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.
• Oct 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 ...

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

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

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

square again ...

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

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

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

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

$\displaystyle x = 7$ or $\displaystyle x = -1$

be sure to check the validity of both values in the original equation to rule out extraneous solutions.
• Oct 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)
• Oct 13th 2009, 04:54 PM
pacman
see the graph, x = -1 is the only solution.
• Oct 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.
• Oct 14th 2009, 04:34 AM
HallsofIvy
I would separate the two square roots:
$\displaystyle \sqrt{x+2} + \sqrt{3x+4} = 2$
$\displaystyle \sqrt{x+2}= 2- \sqrt{3x+4}$

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

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

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