How do I go about solving this? 2|3x+4y-2|+3sqrt[25-5x+2y]=0

[Kulla_9289]

How do I go about solving this? Here are my steps, but this did not lead anywhere as there are two unknowns. Thanks.

 

[Dr.Peterson]

How do I go about solving this? Here are my steps, but this did not lead anywhere as there are two unknowns. Thanks.

You're right, that sort of work can't lead anywhere; the graph of such an equation will in general be a curve of infinitely many points. There must be something special about it, if they ask for a solution.

Look carefully at the equation: [math]2|3x+4y-2|+3\sqrt{25-5x+2y}=0[/math]
Do you see that the left-hand side can never be negative?

When can it be zero?

 

[Kulla_9289]

Why can't it be negative? Is it because of the modulus and square root if squared? Please don't mind me asking. I am just trying to learn it earnestly

 

[Dr.Peterson]

Why can't it be negative? Is it because of the modulus and square root if squared? Please don't mind me asking. I am just trying to learn it earnestly

It's not about "if squared"! It can't be negative in the first place.

Can the first term (a modulus) ever be negative?

Can the second term (a square root) ever be negative?

Can the sum of two non-negative numbers ever be negative?

What about being zero?

 

[Kulla_9289]

Right. How do I make it zero?

 

[BigBeachBanana]

Right. How do I make it zero?

For [imath]a,b \ge 0[/imath], [imath]a+b = 0 \iff a=0 ~\& ~ b=0[/imath]

This gives 2 equations and 2 unknowns.

 

[Dr.Peterson]

Right. How do I make it zero?

By thinking!

Suppose I said I have two numbers, each of which was either positive or zero, and their sum is zero. What does that tell you about my numbers?

Asking people to do all the thinking for you is not a good way to learn to solve problems yourself ...

 

[Kulla_9289]

That means both of the numbers would be zero. This was my first idea after my first work got me nowhere, but I never proceeded with it because I thought it was too wild.

 

[Mr. Bland]

That means both of the numbers would be zero.

The equation as stated:

[imath]2|3x+4y-2|+3\sqrt{25-5x+2y}=0[/imath]​

If [imath]x=0[/imath] and [imath]y=0[/imath]:

[imath]2|3(0)+4(0)-2|+3\sqrt{25-5(0)+2(0)}[/imath]​

[imath]2|-2|+3\sqrt{25}[/imath]​

[imath]4+15 = 19[/imath]​

There are in fact non-zero values for [imath]x[/imath] and [imath]y[/imath] that satisfy this equation, and that produce the following:

[imath]2|0| + 3\sqrt{0} = 0[/imath]​

Can the second term (a square root) ever be negative?

If we're being pedantic, and considering that I am a pedant...

Let us say that [imath]x=\frac{1673}{416}[/imath] and [imath]y=-\frac{3953}{1664}[/imath]:

[imath]2|3(\frac{1673}{416})+4(-\frac{3953}{1664})-2|+3\sqrt{25-5(\frac{1673}{416})+2(-\frac{3953}{1664})}=0[/imath]​

[imath]2|\frac{5019}{416}-\frac{3953}{416}-2|+3\sqrt{25-\frac{16730}{832}-\frac{3953}{832}}=0[/imath]​

[imath]2|\frac{9}{16}|+3\sqrt{\frac{9}{64}}=0[/imath]​

[imath]\frac{9}{8}+3(-\frac{3}{8})=0[/imath]​

@Kulla_9289 this is very much not the answer your assignment wants you to use.

 

[Dr.Peterson]

That means both of the numbers would be zero. This was my first idea after my first work got me nowhere, but I never proceeded with it because I thought it was too wild.

Presumably you didn't mean x=y=0, but that the two terms of the equation are zero, which is where I was leading you.

So do it! Never stop just because something feels new to you, or even "wild", unless that means "wrong". Always try. If the answer you get fails a check, then you can go back and try something else, or just pause to think (as in fact you did, when the work you did led nowhere).

If we're being pedantic, and considering that I am a pedant...

Not pedantic, just disregarding context -- namely, the fact that the equation is about the (principal) root, which is what the symbol means, and therefore that's what I was referring to: "a [principal] square root", not "either of the two roots".

[imath]2|\frac{9}{16}|+3\sqrt{\frac{9}{64}}=0[/imath]

[imath]\frac{9}{8}+3(-\frac{3}{8})=0[/imath]​

This is just wrong.

 

[Harry_the_cat]

What 2 expressions must be equal to zero?
Write them down and then solve simultaneously.

 

[Kulla_9289]

Here is my working to the final answer:

 

[Steven G]

Why can't it be negative? Is it because of the modulus and square root if squared? Please don't mind me asking. I am just trying to learn it earnestly

2|3x+4y-2|+3sqrt[25-5x+2y]=0​

The left side can't be negative because it equals the right hand side, which is 0. And 0 is not negative.

Also, 2|3x+4y-2| is Not negative and 3sqrt(25-5x+2y) is Not negative. The sum of two non-negative terms will be non-negative.

Suppose x^2 + sqrt(x+2y^2) = 11.
Can you tell us EXACTLY what x^2 + sqrt(x+2y^2) equals?

 

[Steven G]

If 3(anything) = 0, then anything =0. No need to distribute that 3^2.

Your answers look correct. Why not plug your answers into the original problem and see if it works?

 

[Kulla_9289]

If 3(anything) = 0, then anything =0. No need to distribute that 3^2.

Ah yes. I have forgotten about it. I have rewritten my work.

 

[Kulla_9289]

Can you tell us EXACTLY what x^2 + sqrt(x+2y^2) equals?

It will equal to a non-negative number because the sum of two non-negative terms will be non-negative. Squaring is always non-negative and square root cannot be a negative.

 

[Dr.Peterson]

Here is my working to the final answer:

That looks good, though you can save a lot of work.

I just observed that the first term is zero when [imath]3x+4y-2=0[/imath], and the second term is zero when [imath]25-5x+2y=0[/imath]. This gives you a simple system of two equations in two unknowns, with reasonably small numbers. I used the addition method, rather than your substitution method, but that's mostly a matter of taste (and a preference to avoid fractions when possible).

The left side can't be negative because it equals the right hand side, which is 0. And 0 is not negative.

That certainly is not what I meant when I said that. The point is that the left side, in itself, is always positive or zero, and the reason for that (that this is true of each term) leads to the observation that both must be zero in order for the sum to be zero.

 

[Harry_the_cat]

Here is my working to the final answer:

Your answers are correct, but, as Steven G said, you can check it yourself by substituting into the original.

You could have made is easier for yourself by just pulling out
3x + 4y - 2 = 0 and
25 - 5x + 2y = 0
and starting there.

EDIT: What Dr P said!

 

[Kulla_9289]

Oh yes. How have I missed that. I have rewritten it.

 

[Steven G]

It will equal to a non-negative number because the sum of two non-negative terms will be non-negative. Squaring is always non-negative and square root cannot be a negative.

No, no no.
If x^2 + sqrt(x+2y^2) = 11, then the exact value of x^2 + sqrt(x+2y^2) is 11!!

Do you see that??

Just like when you are given 2|3x+4y−2|+ sqrt(25−5x+2y)=0, it follows that 2|3x+4y−2|+ sqrt(25−5x+2y) is 0.

Now if you were just given 2|3x+4y−2|+ sqrt(25−5x+2y), then you can say that it is non-negative.