Solve the following equation for x

[WafflesPancakes]

[MATH]x + \sqrt{x+3} = 3 + 3\sqrt{1-x}[/MATH]

 

[lev888]

There is a square root of 1? Doesn't look right to me.

 

[lex]

Continuing the theme from a recent thread!

Spoiler

[MATH]x+\sqrt{x+3}=3+3\sqrt{1-x}\hspace2ex \text{(1)}\\ \text{ }\\ \sqrt{1-x} \text{ means }\boxed{\; x≤1\;}\\ \text{Therefore }\sqrt{x+3}≤2\\ x+\sqrt{x+3}≤x+2\\ \text{i.e. }3+3\sqrt{1-x}≤x+2\hspace2ex \text{ by (1)}\\ x≥1+3\sqrt{1-x}≥1\\ \text{i.e. } \boxed{\;x≥1\;}[/MATH]So [MATH]x=1[/MATH] is the only possible solution.
Now to check whether it is a solution, substitute it into equation (1) and it clearly works.
Solution: [MATH]x=1[/MATH]

 

[HallsofIvy]

Get rid of the square roots by squaring!

\(\displaystyle \sqrt{x+ 3}= 3- x+ \sqrt{1- x}\)
\(\displaystyle x+ 3= (3- x)^2+ 2(3- x)\sqrt{1- x}+ 1- x= 10- 7x+ x^2+ (6-2x)\sqrt{1- x}\)
\(\displaystyle x^2- 8x+ 7= (2z- 6)\sqrt{1- x}\)

Square again:
\(\displaystyle x^4- 9x^3+ 7x^2- 8x^3+ 64x^2- 56x+ 7x^2- 56x+ 49= (4x^2- 24x+ 36)(1- x)= 4x^2- 24x+ 36- 4x^3+ 24^2- 36x\)
\(\displaystyle x^4- 17x^3+ 78x^2- 112x+ 49= -4x^3+ 28x^2-60x+ 36\)
\(\displaystyle x^4- 13x^3+ 50x^2- 52x+ 13= 0\).

 

[Deleted member 4993]

Get rid of the square roots by squaring!

\(\displaystyle \sqrt{x+ 3}= 3- x+ \sqrt{1- x}\)
\(\displaystyle x+ 3= (3- x)^2+ 2(3- x)\sqrt{1- x}+ 1- x= 10- 7x+ x^2+ (6-2x)\sqrt{1- x}\)
\(\displaystyle x^2- 8x+ 7= (2z- 6)\sqrt{1- x}\)

Square again:
\(\displaystyle x^4- 9x^3+ 7x^2- 8x^3+ 64x^2- 56x+ 7x^2- 56x+ 49= (4x^2- 24x+ 36)(1- x)= 4x^2- 24x+ 36- 4x^3+ 24^2- 36x\)
\(\displaystyle x^4- 17x^3+ 78x^2- 112x+ 49= -4x^3+ 28x^2-60x+ 36\)
\(\displaystyle x^4- 13x^3+ 50x^2- 52x+ 13= 0\).

@HoI

There are typos in the response above. In addition there is a mistake somewhere (because x=1 should be a root). If you post a corrected version - I'll delete the response above.

 

[WafflesPancakes]

TYSM for the solution, its really elegant

 

[WafflesPancakes]

Get rid of the square roots by squaring!

\(\displaystyle \sqrt{x+ 3}= 3- x+ \sqrt{1- x}\)
\(\displaystyle x+ 3= (3- x)^2+ 2(3- x)\sqrt{1- x}+ 1- x= 10- 7x+ x^2+ (6-2x)\sqrt{1- x}\)
\(\displaystyle x^2- 8x+ 7= (2z- 6)\sqrt{1- x}\)

Square again:
\(\displaystyle x^4- 9x^3+ 7x^2- 8x^3+ 64x^2- 56x+ 7x^2- 56x+ 49= (4x^2- 24x+ 36)(1- x)= 4x^2- 24x+ 36- 4x^3+ 24^2- 36x\)
\(\displaystyle x^4- 17x^3+ 78x^2- 112x+ 49= -4x^3+ 28x^2-60x+ 36\)
\(\displaystyle x^4- 13x^3+ 50x^2- 52x+ 13= 0\).

Not the most elegant way of doing so, especially for a non-calculator test. But thanks anyways

 

[Cubist]

Continuing the theme from a recent thread...

Absolutely brilliant post! Can I ask @lex if are there any rules (or things to spot) before using this type of method? The following change to the original question gives a less useful result...

[MATH]x+\sqrt{x+3}=3+3\sqrt{\color{red}2\color{black}-x}\hspace2ex \text{(1)}\\ \text{ }\\ \sqrt{2-x} \text{ means }\boxed{\; x≤2\;}\\ \text{Therefore }\sqrt{x+3}≤\sqrt{5}\\ x+\sqrt{x+3}≤x+\sqrt{5}\\ \text{i.e. }3+3\sqrt{2-x}≤x+\sqrt{5}\hspace2ex \text{ by (1)}\\ x≥3-\sqrt{5}+3\sqrt{2-x}≥3-\sqrt{5}\\ \boxed{x≥3-\sqrt{5}}[/MATH]Approximately then 0.76 < x ≤ 2. Is there a substitution or similar that can tighten this up (to become an equivalence), or must we resort to the squaring technique (and then subsequent elimination of false, introduced, results) in this case?

By the way, here's the other thread (I think!)

 

[Cubist]

Here's a way of doing the original question by squaring...

[math]x + \sqrt{x+3} = 3 + 3\sqrt{1-x}[/math]
Make the following substitution to "remove" a sqrt

[math]u = \sqrt{1-x} \implies u^2 = 1-x \implies x=1-u^2[/math]
[math]1 - u^2 + \sqrt{4-u^2} = 3 + 3u\\ \sqrt{4-u^2} = u^2 + 3u + 2\\ \sqrt{(2-u)(u+2)} = (u + 2)(u + 1)[/math]Then square EDIT: I initially tried this without factoring on the line above, and that led to a quite a mess!

[math](2 - u)\bcancel{(u + 2)} = (u + 2)^{\bcancel{2}}(u + 1)^2[/math]
u=-2 is a solution

[math](2 - u) = (u + 2)(u + 1)^2\\ 2 - u = u^3 + 4u^2 + 5u + 2\\ 0 = u^3 + 4u^2 + 6u\\ 0 = \bcancel{u} (u^2 + 4u + 6)[/math]u=0 is a solution. Getting back to x values...

[math]u=0 \implies x=1\\ u=-2 \implies x=-3[/math]Plugging back into the original reveals that x=1 is the only valid solution of the two.

 

[lex]

@Cubist
Nice work.
I'm afraid for my method you cannot be sure that you are going to narrow it down to one potential solution. There was a large element of luck just.

 

[Deleted member 4993]

@Cubist
Nice work.
I'm afraid for my method you cannot be sure that you are going to narrow it down to one potential solution. There was a large element of luck just.

Only goes to support that - practice makes you lucky.....

 

[WafflesPancakes]

@Cubist
Nice work.
I'm afraid for my method you cannot be sure that you are going to narrow it down to one potential solution. There was a large element of luck just.

Well if you have something that you could block the value in, both ways then i guess you could do it?! Maybe...

 

[lex]

@Subhotosh Khan
It's been proven again!

 

[Cubist]

I'm afraid for my method you cannot be sure that you are going to narrow it down to one potential solution. There was a large element of luck just.

With a little practice a good student could probably learn to quickly spot if this shortcut can be applied.
It does seem very unlikely that this kind of question would appear in a test unless there's a trick.

Only goes to support that - practice makes you lucky.....

True that! I too got lucky since there was no guarantee that I wouldn't end up with a fourth degree poly in post#9. My chances of spotting the common factor just before squaring were improved due to past practice.

 

[Steven G]

Here's a way of doing the original question by squaring...

[math]x + \sqrt{x+3} = 3 + 3\sqrt{1-x}[/math]
Make the following substitution to "remove" a sqrt

[math]u = \sqrt{1-x} \implies u^2 = 1-x \implies x=1-u^2[/math]
[math]1 - u^2 + \sqrt{4-u^2} = 3 + 3u\\ \sqrt{4-u^2} = u^2 + 3u + 2\\ \sqrt{(2-u)(u+2)} = (u + 2)(u + 1)[/math]Then square EDIT: I initially tried this without factoring on the line above, and that led to a quite a mess!

[math](2 - u)\bcancel{(u + 2)} = (u + 2)^{\bcancel{2}}(u + 1)^2[/math]
u=-2 is a solution

[math](2 - u) = (u + 2)(u + 1)^2\\ 2 - u = u^3 + 4u^2 + 5u + 2\\ 0 = u^3 + 4u^2 + 6u\\ 0 = \bcancel{u} (u^2 + 4u + 6)[/math]u=0 is a solution. Getting back to x values...

[math]u=0 \implies x=1\\ u=-2 \implies x=-3[/math]Plugging back into the original reveals that x=1 is the only valid solution of the two.

u=-2 is a (possible) solution? Where did that come from or was it a small error on your part?

 

[lookagain]

Here are some main (and lesser) steps of a partial solution that is similar
in manner to post # 9.

Let \(\displaystyle \ x = u^2 - 3, \ \ u \ge 0.\)

. . .

\(\displaystyle u^2 + u - 6 = 3\sqrt{4 - u^2} \)

. . .

\(\displaystyle (u + 3)^2(u - 2)^2 = 9(4 - u^2) \)

\(\displaystyle (u + 3)^2(u - 2)^2 = -9(u^2 - 4)\)

\(\displaystyle (u + 3)^2(u - 2)^2 + 9(u - 2)(u + 2) = 0 \)

\(\displaystyle (u - 2)[(u + 3)^2(u - 2) + 9(u + 2)] = 0 \)

. . .

\(\displaystyle (u - 2)(u^3 + 6u^2 - 2u^2 - 12u + 9u - 18 + 9u + 18) = 0\)

\(\displaystyle (u - 2)(u^3 + 4u^2 + 6u) = 0 \)

\(\displaystyle u(u - 2)(u^2 + 4u + 6) = 0\)

\(\displaystyle u = 0 \implies x = -3\)

\(\displaystyle u - 2 = 0 \implies u = 2 \implies x = 1\)

\(\displaystyle u^2 + 4u + 6 = 0 \ \ has \ \ no \ \ real \ \ solutions. \)

x = 1 is the only value that checks in the original equation.


------------------------------------------------------------------------------------------------


Edit: \(\displaystyle \ \ \) Jomo, look how u = -2 makes each side zero in Cubist's
u-variable equation right above that.

Also, Cubist's equation where u = -2 as a candidate is equivalent to:

\(\displaystyle (2 - u)(u + 2) - (u + 2)^2(u + 1)^2 = 0 \)

 

[Cubist]

Here are some main (and lesser) steps of a partial solution that is similar
in manner to post # 9...

Nice! Proving that it works when a substitution is made to eliminate the other sqrt!

\(\displaystyle \ \ \) Jomo, look how u = -2 makes each side zero in Cubist's
u-variable equation right above that.

Also, Cubist's equation where u = -2 as a candidate is equivalent to:

\(\displaystyle (2 - u)(u + 2) - (u + 2)^2(u + 1)^2 = 0 \)

Yes indeed! @Jomo I was just being lazy because I didn't want to write (u+2) as a factor on all the subsequent lines. But I admit that it wasn't very clear. It's probably a bad habit of mine that has leaked out!

Well if you have something that you could block the value in, both ways then i guess you could do it?! Maybe...

I can't think of a way. As far as I can see, IF I've understood correctly, the solution in post#3 is mostly useful when:-

• you can spot a solution by inspection ( maybe luck ? ) OR a solution is provided
• AND the solution is at the end of the domain for one (or both) side(s) of the equation
• AND your goal is to prove the solution is valid and unique

See the following graph of the LHS(red) and RHS(blue) of the original equation...


Lex takes advantage that the curves meet(touch) at end of blue's domain, and then effectively proves that the red line is always below the blue line when x<1 AND that they meet at x=1.

In fact that technique could be used if you want to prove a known solution that isn't at the end of a domain. You'd just have to prove x in the region to the left, and once more for the domain to the right of the solution.

I personally recommend that you don't spend too long studying the post#3 method (unless you're comfortable with it and fully understand). The squaring technique is more "generally useful" since it works mid-domain and also when you don't know the solution in advance. The trick to performing it quickly is to spot methods of getting/ keeping the poly in a factored form (not always possible in real life, but with questions on a test there's usually a way ). If you spot an exact solution you can also use this information to factor the poly. This advice is NOT intended to detract from @lex's post which is brilliant, quick and a very clever spot!

 

[lex]

Another little digression.

You can get potential solutions, by iterating using your calculator, rewriting the equation to get x=...
E.g. Consider

[MATH]\text{1. }\hspace2ex 2\sqrt{1-x}+1=\sqrt{2x+1}[/MATH]
Rearranging
[MATH]\sqrt{1-x}=\dfrac{\sqrt{2x+1}-1}{2}[/MATH][MATH]x=1-\left(\dfrac{\sqrt{2x+1}-1}{2}\right)^2[/MATH]Type this in your calculator (Casio fx83, cheap and cheerful) and then all you have to do is keep pressing equals:

(I started the process by typing "2="; that puts '2' into Ans, to get started).
I get the potential solution 8/9, which I can see is a solution, by substituting into the original equation.

Some example equations. (1. is as above).

[MATH]\text{1. }\hspace2ex 2\sqrt{1-x}+1=\sqrt{2x+1}\\ \text{2. }\hspace2ex x+\sqrt{x+3}=3+3\sqrt{1-x}\\ \text{3. }\hspace2ex x+\sqrt{x+3}=3+3\sqrt{2-x}\\ \text{4. }\hspace2ex \sqrt{2x-1}=1-\sqrt{x-1}\\ \text{5. }\hspace2ex \sqrt{1-2x}=1-\sqrt{x+1}[/MATH]
Potential rearrangements:
1. Done
2. [MATH]x=1-\left(\frac{x+\sqrt{x+3}-3}{3}\right)^2[/MATH]
3. [MATH]x=2-\left(\frac{x+\sqrt{x+3}-3}{3}\right)^2[/MATH]4. [MATH]x=(1-\sqrt{2x-1})^2+1[/MATH]5. [MATH]x=(1-\sqrt{1-2x})^2-1[/MATH]

 

[HallsofIvy]

I am so glad to know that the Casio fx83 is not only cheap but cheerful!

I do love a cheerful calculator!

 

[lex]

@HallsofIvy
Glad to see a fellow cheerful calculator.