Solving Radical Equations sqrt{x^2 - 2x + 1} = 2x + 5

[salk26]

(Solve the following radical equations. Be sure to identify any extraneous solutions.)
sqrt{x^2 - 2x + 1} = 2x + 5
I’m supposed to get -4/3 but got -6. The way I did it is the inside part of the square root can be factored to (x-1)^2 so I just got rid of the square root without having to square both sides of the equation. Then it’s x-1=2x+5 which is solved to be x=-6. But if I plug that back into the original equation it’s not possible so there’s no solutions. So where did I go wrong where I was supposed to get -4/3?

 

[Aion]

Since [imath]\sqrt{(x-1)^2}= \vert x-1 \vert =2x+5[/imath]. Are you familiar with the absolute value function? When you take the square root of a perfect square, you must consider two possible cases. For suppose [imath]{(x-1)^2}=\Delta[/imath]. In the next step we must consider [imath]{\delta}^2=\Delta[/imath]. Then the equation becomes [imath](x-1)^2=\delta^2[/imath]. By taking the square root we obtain two solutions. Namely [imath]x-1=\delta[/imath] or [imath]x-1=-\delta[/imath]. The reason for this is because [imath](-{\delta})^2=\delta^2[/imath].

 

[Aion]

Since [imath]\sqrt{(x-1)^2}= \vert x-1 \vert =2x+5[/imath]. Are you familiar with the absolute value function? When you take the square root of a perfect square, you must consider two possible cases. For suppose [imath]{(x-1)^2}=\Delta[/imath]. In the next step we must consider [imath]{\delta}^2=\Delta[/imath]. Then the equation becomes [imath](x-1)^2=\delta^2[/imath]. By taking the square root we obtain two solutions. Namely [imath]x-1=\delta[/imath] or [imath]x-1=-\delta[/imath]. The reason for this is because [imath](-{\delta})^2=\delta^2[/imath].

To be more pedagogical, consider that the roots from the equation above can also be written as


[math]\delta =x-1 \quad \text{or} \quad \delta=-(x-1)[/math]
since [imath](-(x-1))^2=(x-1)^2[/imath].

 

[mario99]

Since [imath]\sqrt{(x-1)^2}= \vert x-1 \vert =2x+5[/imath]. Are you familiar with the absolute value function? When you take the square root of a perfect square, you must consider two possible cases. For suppose [imath]{(x-1)^2}=\Delta[/imath]. In the next step we must consider [imath]{\delta}^2=\Delta[/imath]. Then the equation becomes [imath](x-1)^2=\delta^2[/imath]. By taking the square root we obtain two solutions. Namely [imath]x-1=\delta[/imath] or [imath]x-1=-\delta[/imath]. The reason for this is because [imath](-{\delta})^2=\delta^2[/imath].

Aion, do you know that this is one of the most interesting problems because it involves the confusing absolute value. In my opinion, if this problem was given in a test, [imath]99[/imath]% of the students would fail to see [imath]\sqrt{x^2 - 2x + 1} = |x - 1|[/imath]. And [imath]99[/imath]% of the remaining [imath]1[/imath]% would fail to solve [imath]|x-1| = 2x + 5[/imath].

 

[Dr.Peterson]

Aion, do you know that this is one of the most interesting problems because it involves the confusing absolute value. In my opinion, if this problem was given in a test, [imath]99[/imath]% of the students would fail to see [imath]\sqrt{x^2 - 2x + 1} = |x - 1|[/imath]. And [imath]99[/imath]% of the remaining [imath]1[/imath]% would fail to solve [imath]|x-1| = 2x + 5[/imath].

On the other hand, many students, not thinking quite so deeply, might not notice the perfect square, and would just follow the usual routine of squaring the equation [imath]\sqrt{x^2 - 2x + 1} = 2x + 5[/imath] to get
[math]x^2 - 2x + 1 = (2x + 5)^2\\x^2 - 2x + 1 = 4x^2+20x+25\\3x^2+22x+24=0\\(3x+4)(x+6)=0[/math]and conclude that [imath]x=-\frac{4}{3}[/imath] or [imath]x=-6[/imath].

They would check these and find that the latter is extraneous, but the former works. So they would get it right because they didn't notice the shortcut!

Lesson: When you are smart enough to find a quick way to do something, use the time you save to think very carefully, rather than running so fast you trip.

 

[khansaheb]

solved to be x=-6.

Incorrect .... please show detailed computation.

 

[mario99]

On the other hand, many students, not thinking quite so deeply, might not notice the perfect square, and would just follow the usual routine of squaring the equation [imath]\sqrt{x^2 - 2x + 1} = 2x + 5[/imath] to get
[math]x^2 - 2x + 1 = (2x + 5)^2\\x^2 - 2x + 1 = 4x^2+20x+25\\3x^2+22x+24=0\\(3x+4)(x+6)=0[/math]and conclude that [imath]x=-\frac{4}{3}[/imath] or [imath]x=-6[/imath].

They would check these and find that the latter is extraneous, but the former works. So they would get it right because they didn't notice the shortcut!

Lesson: When you are smart enough to find a quick way to do something, use the time you save to think very carefully, rather than running so fast you trip.

I am against you Dr.Peterson in this part. I prefer to solve the problem quickly to get:

[imath](3x+4)(x+6)=0[/imath]

[imath]\displaystyle x=-\frac{4}{3}[/imath] or [imath]x=-6[/imath].

And then checking the validity of the solutions, rather than being smart and wasting the time.

But this preferable method has some flaw. Most of the students don't check their solutions. I was one of them. Therefore, if the problem did not alert them by "Be sure to identify any extraneous solutions", they would fail to get the full mark.

 

[Dr.Peterson]

I am against you Dr.Peterson in this part

Huh? It sounds like you're agreeing with me. There are two methods, and each has a flaw; I don't prefer either one.

The routine way is slow and sure (as long as you don't forget to check for extraneous solutions), while the other is quick and risky (because it's easy to miss the absolute value). Are you assuming that what I say the smart person would do, I am saying is the smarter way? As I tell students, the best way is the way you see!

 

[Dr.Peterson]

Incorrect .... please show detailed computation.

But they did:

The way I did it is the inside part of the square root can be factored to (x-1)^2 so I just got rid of the square root without having to square both sides of the equation. Then it’s x-1=2x+5 which is solved to be x=-6.

And that's correct, as far as they went, forgetting the absolute value. They even checked it and saw that it didn't work.

So almost everything was done quite well. It's just that the one mistake lost the correct solution.

 

[mario99]

Huh? It sounds like you're agreeing with me. There are two methods, and each has a flaw; I don't prefer either one.

The routine way is slow and sure (as long as you don't forget to check for extraneous solutions), while the other is quick and risky (because it's easy to miss the absolute value). Are you assuming that what I say the smart person would do, I am saying is the smarter way? As I tell students, the best way is the way you see!

At first, I thought that you meant you preferred the smarter way, but now realized that:

Lesson: When you are smart enough to find a quick way to do something, use the time you save to think very carefully, rather than running so fast you trip.

The sentence above meant: even the smarter way may cost you some credits if you are not very careful. I got your point.