Solving Equations Homework: Question 4(c) Part I & II

[halvizo1031]

Homework Statement

I am having trouble with question 4(c)... part i and ii...

Homework Equations

The Attempt at a Solution

would it be okay for her to write iff between each line? I do not see why not but I cannot find the proper reasoning.

 

[micromass]

Well, you're correct. In her attempt in part (i), it is ok to write iff in every line. Can you argue for each line why it is allowed?

 

[halvizo1031]

the only reason i can think of is that she's not really changing the orginal equation. she is only simplying it.

 

[micromass]

Yes! If I were your instructor, that would certainly be a good answer. But I don't really know what they want to hear. You could also mention the laws involved: for example, you could say that in the first line, she used distributivity, etc.

Anyway, what do you think of the second part?

 

[halvizo1031]

well my thinking for part ii is that since (x-3) is always smaller than (x), then we would have a small number minus a big number which would give us a negative solution...thus, since the equation is equalled to 5, there must be no solution.

 

[micromass]

Yes, but it would be nice to elaborate a bit.

For example, x-3\leq x, implies that \sqrt{x-3}\leq \sqrt{x}. thus \sqrt{x-3}-\sqrt{x}\leq 0, which implies that the left-hand side can never equal 5.

 

[halvizo1031]

i see...then, may I use the same reasoning for part (d)?

 

[micromass]

Can you explain your reasoning?? As the question states, you will not be able to use the exact same reasoning here!

 

[halvizo1031]

i need some time to think about the reasoning here...i'm not quite sure i see it.

 

[micromass]

remember that a square root is only defined if the argument is positive...

 

[halvizo1031]

thus, sqrt of 1-x will always give us a complex answer?

 

[micromass]

Not necessarily, since x could be 0 for example, then \sqrt{1-x}=1. But I like your way of thinking...

 

[halvizo1031]

i'm not quite sure what it is the question is looking for ...

 

[halvizo1031]

it seems to me that stating that the equation will give a complex answer when x>1

 

[Char. Limit]

it seems to me that stating that the equation will give a complex answer when x>1

Yes, when x>1, then sqrt(1-x) is not a real number. Now, what about your other term, sqrt(x-1)? What is that when x<1?

 

[halvizo1031]

if x < 1 then i will also get a complex solution here...therefore, i will always have a complex solution to the equation for any value of x except 0?

 

[micromass]

Indeed, so can the left-hand side ever equal 1?

 

[hunt_mat]

My only concern is that in the third line she uses:

<br /> \sqrt{a^{2}}=a<br />
and not:

<br /> \sqrt{a^{2}}=\pm a<br />

 

[micromass]

But he simply does \sqrt{4}=2, which is certainly correct? The square root is always a positive number...

 

[hunt_mat]

Perhaps, but I was a little unsure about that. I don't know what the conventions they're using.

 

[Char. Limit]

The usual convention in mathematics is that this sign: \sqrt{} is used to define the positive square root of a number.

 

[halvizo1031]

the way i see it is that for any value i plug in for x (except 1), the left hand side will always give me a complex answer which does not equal 1.

 

[micromass]

the way i see it is that for any value i plug in for x (except 1), the left hand side will always give me a complex answer which does not equal 1.

Yes, so the only possible solution is 1, which can be easily checked.

 

[halvizo1031]

hold on, if i plug in 1 for x, wouldn't i get 0 = 1?

 

[Char. Limit]

hold on, if i plus in 1 for x, wouldn't i get 0 = 1?

And because that's not true, you've proven that it cannot be true for x<1, x>1, or x=1. Congrats!

 

[halvizo1031]

thanks! i appreciate everyone's help.