Quadratic Form Rational Equation - Solving Without U-Substitution

[spenc193]

2/y - 1/y^1/2 + 1 = 0

I know that this equation can be solved with u-substitution, where u = y^-1/2, but I was attempting to solve this by multiplying through instead with a LCD.

So I:

(1) Multiplied through by 'y', giving me:

2 - y^3/2 + y = 0

then,

(2) Squared the entire equation, giving:

4 + y³ + y² = 0


I know this is probably wrong to start with, and even if it isn't, I do not know where to proceed. Any help would be really appreciated. Thanks guys

 

[Deleted member 4993]

2/y - 1/y^1/2 + 1 = 0

I know that this equation can be solved with u-substitution, where u = y^-1/2, but I was attempting to solve this by multiplying through instead with a LCD.

So I:

(1) Multiplied through by 'y', giving me:

2 - y^3/2 + y = 0

then,

(2) Squared the entire equation, giving:

4 + y³ + y² = 0


I know this is probably wrong to start with, and even if it isn't, I do not know where to proceed. Any help would be really appreciated. Thanks guys

(2) Squared the entire equation, giving:

4 + y³ + y² = 0

This step above is incorrect.

(a + b + c)² = a² + b² + c² + 2*a*b + 2*b*c + 2*c*a

You are missing some terms.

Is this

(1) a class-assignment or

(2) you are trying something out of curiosity

 

[spenc193]

I see this mistake now, yes. A silly one too. I am doing this purely out of curiosity. It was a class problem meant to be solved with U-sub but I wanted to wrestle with it without that.

I am thinking, then, that it would probably be easier to isolate the fractional exponent on one side of the argument and then proceed with squaring. Or, do the trinomial squaring you showed.

What if I did the following (beginning at my previous step [1])?

y - y^3/2 + 2 = 0

[isolate frac exponent]

-y^3/2 = -y - 2

[multiply through by -1]

y^3/2 = y + 2

[square the equation]

y³ = y² -4y + 4

[standardize]

y³ - y² + 4y - 4 = 0


Now here's the only problem, I don't know how to solve/factor cubic equations yet! Thank you for the help. If you have any tips on solving cubics I'd be open and grateful.

 

[MarkFL]

If you multiply the original equation by \(y\), then you should have:

[MATH]2-y^{\frac{1}{2}}+y=0[/MATH]
Or:

[MATH]y-y^{\frac{1}{2}}+2=0[/MATH]
Then:

[MATH]y+2=y^{\frac{1}{2}}[/MATH]
Square:

[MATH]y^2+4y+4=y[/MATH]
[MATH]y^2+3y+4=0[/MATH]
Now, apply the quadratic formula...

 

[Deleted member 4993]

I see this mistake now, yes. A silly one too. I am doing this purely out of curiosity. It was a class problem meant to be solved with U-sub but I wanted to wrestle with it without that.

I am thinking, then, that it would probably be easier to isolate the fractional exponent on one side of the argument and then proceed with squaring. Or, do the trinomial squaring you showed.

What if I did the following (beginning at my previous step [1])?

y - y^3/2 + 2 = 0

[isolate frac exponent]

-y^3/2 = -y - 2

[multiply through by -1]

y^3/2 = y + 2

[square the equation]

y³ = y² -4y + 4 ................... incorrect - that should be + 4 * y

[standardize]

y³ - y² + 4y - 4 = 0


Now here's the only problem, I don't know how to solve/factor cubic equations yet! Thank you for the help. If you have any tips on solving cubics I'd be open and grateful.

There is a method called Cardan's method to solve cubic equation. However it is too messy and I have never used it. At this point I generally go for "numerical" approximations (good enough for government work).

 

[spenc193]

There is a method called Cardan's method to solve cubic equation. However it is too messy and I have never used it. At this point I generally go for "numerical" approximations (good enough for government work).

I will check this out. Thanks very much.

 

[JeffM]

2/y - 1/y^1/2 + 1 = 0

I know that this equation can be solved with u-substitution, where u = y^-1/2, but I was attempting to solve this by multiplying through instead with a LCD.

So I:

(1) Multiplied through by 'y', giving me:

2 - y^3/2 + y = 0

then,

(2) Squared the entire equation, giving:

4 + y³ + y² = 0


I know this is probably wrong to start with, and even if it isn't, I do not know where to proceed. Any help would be really appreciated. Thanks guys

Aside from the fact that you squared incorrectly, as Subhotosh pointed out, you made things hard for yourself by not simplifying before squaring.

[MATH]\dfrac{2}{y} - \dfrac{1}{y^{1/2}} + 1 = 0 \implies \dfrac{1}{y^{1/2}} - \dfrac{2}{y} = 1.[/MATH]
Now you have no trinomial to square. And do you remember about common denominators?

[MATH]\therefore \dfrac{1}{y^{1/2}} * \dfrac{y^{1/2}}{y^{1/2}} - \dfrac{2}{y} = 1 \implies \dfrac{y^{1/2}}{y} - \dfrac{2}{y} = 1 \implies \dfrac{y^{1/2} - 2}{y} = 1.[/MATH]
Now clear fractions.

[MATH]\therefore y = y^{1/2} - 2 = \sqrt{y} - 2.[/MATH]
Now the way to avoid fractional exponents is to isolate the square root.

[MATH]\therefore y + 2 = \sqrt{y} \implies (y + 2)^2 = (\sqrt{y})^2 \implies y^2 + 4y + 4 = y.[/MATH]
You were taught how to simplify things for a reason.

 

[spenc193]

If you multiply the original equation by \(y\), then you should have:

[MATH]2-y^{\frac{1}{2}}+y=0[/MATH]

Hey Mark,

Multiplying through by 'y', when you get to that second term, you wind up with y¹/y^-1/2, correct?

That would result in y^3/2?

 

[spenc193]

Aside from the fact that you squared incorrectly, as Subhotosh pointed out, you made things hard for yourself by not simplifying before squaring.

[MATH]\dfrac{2}{y} - \dfrac{1}{y^{1/2}} + 1 = 0 \implies \dfrac{1}{y^{1/2}} - \dfrac{2}{y} = 1.[/MATH]
Now you have no trinomial to square. And do you remember about common denominators?

[MATH]\therefore \dfrac{1}{y^{1/2}} * \dfrac{y^{1/2}}{y^{1/2}} - \dfrac{2}{y} = 1 \implies \dfrac{y^{1/2}}{y} - \dfrac{2}{y} = 1 \implies \dfrac{y^{1/2} - 2}{y} = 1.[/MATH]
Now clear fractions.

[MATH]\therefore y = y^{1/2} - 2 = \sqrt{y} - 2.[/MATH]
Now the way to avoid fractional exponents is isolate the square root.

[MATH]\therefore y + 2 = \sqrt{y} \implies (y + 2)^2 = (\sqrt{y})^2 \implies y^2 + 4y + 4 = y.[/MATH]
You were taught how to simplify things for a reason.

Great explanation. It's appreciated.

 

[MarkFL]

Hey Mark,

Multiplying through by 'y', when you get to that second term, you wind up with y¹/y^-1/2, correct?

That would result in y^3/2?

In your original post, it looks to me like that second term is:

[MATH]\frac{1}{y^{\frac{1}{2}}}[/MATH]
I see no negative on that exponent, as given.

 

[HallsofIvy]

You should be able to see, "by inspection", that y= 1 is a solution to \(\displaystyle y^3- y^2+ 4y- 4= 0\). Dividing \(\displaystyle y^3- y^2+ 4y- 4\) by \(\displaystyle y- 1\) leaves \(\displaystyle y^2+ 4= 0\) that has no real roots but two imaginary roots, i and -i.

 

[spenc193]

In your original post, it looks to me like that second term is:

[MATH]\frac{1}{y^{\frac{1}{2}}}[/MATH]
I see no negative on that exponent, as given.

Okay, I see where part of my confusion was. In my book, the second term is given as 1 over the 'y', where the 'y' is beneath a radical. I assumed that the exponent form of a radical in a denominator was negative because it was in the denominator.

But I suppose that is only if it isn't written in rational form...i.e.

y^-1/2 = 1/y^1/2

 

[spenc193]

Thanks to all posters who responded. It is apparent I should not have wound up with any cubed terms. Apologies to any responders I led that direction, although your tips gave me some things to learn about cubic equations.