Solve √(x^2+3x+7)-√(x^2-3x+9)+2=0

[SigmaS]

I'm struggling with two particular equations in my Algebra textbook.

1) $x^2+m^2=2mx+(nx)^2 $
2) $\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}+2=0$

I know the answer to the first equation is {$\frac{m}{1-n}, \frac{m}{1+n}$}. The answer to the second equation is {$-\frac{9}{5}$}.

I don't understand why I'm struggling on problems that look so simple. Here's how I approached the first problem:
$$x^2+m^2-2mx=n^2(x^2)$$
$$x^2+2mx+m^2-2mx=n^2x^2$$
$$\sqrt{x^2+m^2}=\sqrt{n^2x^2}$$
$$x=\frac{x+m}{n}, \frac{x-m}{n}$$

I suspect my answer is actually correct but it is only translated differently. I'm not too sure. I did plug my answer into a simplified version of the problem and indeed, I got, $x=\frac{x+m}{n}, \frac{x-m}{n}$ but I don't think that's trustworthy.

My approach to the second problem goes as follows:
$$\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}=-2$$
$$(\sqrt{x^2+3x+7})^2-(\sqrt{x^2-3x+9})^2=(-2)^2$$
$$x^2+3x+7-x^2+3x-9=4$$
$$6x=6$$
$$x=1$$

I don't believe squaring the roots actually works the way I tried here, but this was the best I could do. I tried factoring the polynomials within the roots individually, but that led to some really messy results so I avoided it.

Could someone please explain each step of either problem using TeX commands? Other people I asked either gave vague answers without much explanation or did part of the problem in hopes I would connect the dots and complete the rest. I'd really just appreciate a straight forward, detailed solve. Also, I would prefer if you do not skip steps - I don't mind if it's very long.

 

[mathmari]

I don't understand why I'm struggling on problems that look so simple. Here's how I approached the first problem:
$$x^2+m^2-2mx=n^2(x^2)$$
$$x^2+2mx+m^2-2mx=n^2x^2$$
$$\sqrt{x^2+m^2}=\sqrt{n^2x^2}$$
$$x=\frac{x+m}{n}, \frac{x-m}{n}$$

I suspect my answer is actually correct but it is only translated differently. I'm not too sure. I did plug my answer into a simplified version of the problem and indeed, I got, $x=\frac{x+m}{n}, \frac{x-m}{n}$ but I don't think that's trustworthy.

At the second line you added the term 2mx just on the left side of the equation. Instead, we can use the formula $(a-b)^2=a^2-2ab+b^2$. Then we have the following:
\begin{equation*}x^2+m^2-2mx=n^2(x^2) \Rightarrow x^2-2mx+m^2=n^2x^2 \Rightarrow (x-m)^2=(nx)^2 \end{equation*} Now we take the square root and get \begin{equation*}\sqrt{(x-m)^2}=\sqrt{(nx)^2 }\Rightarrow x-m=\pm nx\end{equation*}
So we get \begin{align*}&x-m=nx \ \text{ or } \ x-m=-nx \\ & \Rightarrow x-nx =m \ \text{ or } \ x+nx =m \\ & \Rightarrow x(1-n) =m \ \text{ or } \ x(1+n) =m \\ & \Rightarrow x =\frac{m}{1-n} \ \text{ or } \ x =\frac{m}{1+n}\end{align*}

My approach to the second problem goes as follows:
$$\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}=-2$$
$$(\sqrt{x^2+3x+7})^2-(\sqrt{x^2-3x+9})^2=(-2)^2$$
$$x^2+3x+7-x^2+3x-9=4$$
$$6x=6$$
$$x=1$$

I don't believe squaring the roots actually works the way I tried here, but this was the best I could do. I tried factoring the polynomials within the roots individually, but that led to some really messy results so I avoided it.

At the second line, if you square the equation it should be $(a-b)^2=a^2-2ab+b^2$ on the left side, instead of $a^2-b^2$. Then you have still one square root. You leave this root alone at one side of the equation and square again the equation.

 

[HOI]

The first equation, x^2+ m^2- 2mx= n^2x^2, can be written as (1- n^2)x^2- 2mx+ m^2= 0. Now use the quadratic formula.

For the second equation, \sqrt{x^2+ 3x+ 7}- \sqrt{x^2- 3x+ 9}= 2, no, your "method" is not valid. If a- b= 2 then (a- b)^2= a^2- 2ab+ b^2= 4, not "a^2- b^2= 4".

Because I would really rather not multiply two square roots together my first step would be to rewrite the equation as \sqrt{x^2+ 3x+ 7}= \sqrt{x^2- 3x+ 9}+ 2. NOW square both sides: x^2+ 3x+ 7= x^2- 3x+ 9+ 4\sqrt{x^2- 3x+ 9}+ 4.

Now isolate that remaining square root, 6x- 6= 4\sqrt{x^2- 3x+ 9} or 3x- 3= 2\sqrt{x^2- 3x+ 9}, and square again: 9x^2- 18x+ 9= 4(x^2- 3x+ 9). That is a fairly easy quadratic equation to solve for x.

Caution- when you square both sides of an equation, you may introduce new roots that do not satisfy the original equation. That last quadratic equation will have two roots but they may not both satisfy the original equation. Try both roots in the original equation to see.

 

[Wilmer]

My approach to the second problem goes as follows:
$$\sqrt{x^2+3x+7}-\sqrt{x^2-3x+9}=-2$$
$$(\sqrt{x^2+3x+7})^2-(\sqrt{x^2-3x+9})^2=(-2)^2$$

A simple example to show your method is not quite correct:

sqrt(25) - sqrt(4) = 3

[sqrt(25)]^2 - [sqrt(4)]^2 = 3^2

25 - 4 = 9
21 = 9

Do you "see" now why your method was not quite correct?

 

[SigmaS]

The first equation, x^2+ m^2- 2mx= n^2x^2, can be written as (1- n^2)x^2- 2mx+ m^2= 0. Now use the quadratic formula.

For the second equation, \sqrt{x^2+ 3x+ 7}- \sqrt{x^2- 3x+ 9}= 2, no, your "method" is not valid. If a- b= 2 then (a- b)^2= a^2- 2ab+ b^2= 4, not "a^2- b^2= 4".

Because I would really rather not multiply two square roots together my first step would be to rewrite the equation as \sqrt{x^2+ 3x+ 7}= \sqrt{x^2- 3x+ 9}+ 2. NOW square both sides: x^2+ 3x+ 7= x^2- 3x+ 9+ 4\sqrt{x^2- 3x+ 9}+ 4.

Now isolate that remaining square root, 6x- 6= 4\sqrt{x^2- 3x+ 9} or 3x- 3= 2\sqrt{x^2- 3x+ 9}, and square again: 9x^2- 18x+ 9= 4(x^2- 3x+ 9). That is a fairly easy quadratic equation to solve for x.

Caution- when you square both sides of an equation, you may introduce new roots that do not satisfy the original equation. That last quadratic equation will have two roots but they may not both satisfy the original equation. Try both roots in the original equation to see.

Hey, can you explain why I can rewrite the first equation as (1- n^2)x^2- 2mx+ m^2= 0? It's (1- n^2)x^2 that confuses me. I can complete from there without struggle, but I just don't understand why translating that way works. Also, thank you for the second equation, I finally got the right answer! However, I believe you were supposed to subtract the 2 rather than move it to the other side (in this case it makes no difference because we end up squaring, anyway), like so: \sqrt{x^2+ 3x+ 7}= \sqrt{x^2- 3x+ 9}-2

 

[Wilmer]

Hey, can you explain why I can rewrite the first equation as (1- n^2)x^2- 2mx+ m^2= 0?

x^2 + m^2 - 2mx = n^2 x^2

x^2 - n^2 x^2 + m^2 - 2mx = 0

x^2(1 - n^2) + m^2 - 2mx = 0

Hokay?