# Simple equation.

1. Homework Statement
For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
1) y increases linearly with x
2) y increases as the square of x
3) y increases as y^1/2
4) If x increases, y can increase or decrease.

2. Homework Equations
y is directly proportional to x. as a is a constant, the first option would be right!

3. The Attempt at a Solution

Is my approach right?

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arildno
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Not all, I'm afraid. Rewrite your equation as:
$$(y-(\sqrt{a})x)(y+(\sqrt{a})x)=0$$
Now, two questions:
1. Why is this rewriting permissible?
2. What must you conclude from it?

Last edited:
Not all, I'm afraid. Rewrite your equation as:
$$(y-(\sqrt{a})x)(y+(\sqrt{a})x)=0$$
Now, two questions:
1. Why is this rewriting permissible?
2. What must you conclude from it?
Ans 1 because the difference of two squares can be factorised as suc:
a^2-b^2=(a+b)(a-b)

Ans 2. I dont know what can be concluded! Please be clear!

arildno
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Well, if two numbers multiplied together is zero, what can we say about those numbers?

Well, if two numbers multiplied together is zero, what can we say about those numbers?
That any one of them is equal to zero or both are zero.

arildno
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So,
EITHER:
$$y=\sqrt{a}x$$
OR
$$y=-\sqrt{a}x$$

Given that knowledge, which statement is true?

So,
EITHER:
$$y=\sqrt{a}x$$
OR
$$y=-\sqrt{a}x$$

Given that knowledge, which statement is true?
The fourth option???

arildno
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Yup

but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!!

arildno
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Three flaws:

1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
But that is a totally different issue.

HallsofIvy
Homework Helper
but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!!
This is really only echoing arildno but:

I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.

1. Homework Statement
For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
1) y increases linearly with x
2) y increases as the square of x
3) y increases as y^1/2
4) If x increases, y can increase or decrease.

2. Homework Equations
y is directly proportional to x. as a is a constant, the first option would be right!

3. The Attempt at a Solution

Is my approach right?
Three flaws:

1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
But that is a totally different issue.

If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??

This is really only echoing arildno but:

I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.
Whats the difference between "inearly related" and "increases linearly"?

HallsofIvy
Homework Helper
If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??
Yes, and so with y2= ax2 we can have either
$y= \sqrt{a}x$ or $y= -\sqrt{a}x$

Whats the difference between "inearly related" and "increases linearly"?
In both $y= \sqrt{a}x$ and $y= -\sqrt{a}x$ y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.

Yes, and so with y2= ax2 we can have either
$y= \sqrt{a}x$ or $y= -\sqrt{a}x$

In both $y= \sqrt{a}x$ and $y= -\sqrt{a}x$ y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.
Thanks a lot for the efforts.