As per the given equation: y2+22y+y1/2=−114

By solving this equation we get:

y2+22y+y1/2=-114

y2+y√2+22y=−114

y+22y=-114

23y=-114

y=-114/23

y=-4(22/23)
so the value of y is -4(22/23)

Guest wrote:My guess is we should be able to do......

[tex](a+bi)^2 + 22(a+bi) + (a+bi)^{\frac{1}{2}} = -114[/tex]

And solve a and bi to get an exact answer......

I think we need new ways to solve problems like these. Using current methods I can't find an answer to this problem. This question should probably be asked in the calculus section, it becomes obvious that pure algebra can't solve this, because the constant y never evaluates to -114 in any simple way...

My guess is the answer would be entirely complex.

My guess is we should be able to do......

[tex](a+bi)^2 + 22(a+bi) + (a+bi)^{\frac{1}{2}} = -114[/tex]

And solve a and bi to get an exact answer......

The best answer I can come up with so far is................
y = 0.846 + i13.727
OR
y = 0.846 + i8.294

Oh - where i left off and quit on the quartic formula was here -

was trying rational zeros of polynomials after I used the quartic deformation and simplification, so I wound up with the following after that..

[tex]-8z^3 + (88-24\sqrt{114})z^2 - (1824+176\sqrt{114})z + 1[/tex]
putting in [tex]x = (\frac{p}{q})[/tex] wound up giving me the following
[tex]= \frac{q^3}{8p}[/tex] on the right side
So I took it that would have meant and here is where I don't remember to even folow..
which I thought again was 1/8.

But when I checked it, it didn't work out like it was upposed to, didn't = 0.

So then I usedt he faster way, and that is where I got the othr fractions.

But I have to say, if the equation is workable, it shouldhold true that we can plug in any y value into it, yet it sems to stand that it takes no negative numbers as written, and no imaginary numbers, and no positive numbers work out to -114. So I have to conclude that I was more correct last night in that it doesn't have a completely solvable solution as written...

But then again there's a lot of tricks and techniques I'm not aware of, and we've moved well beyond algebra here...

It wasn't that difficult for me - I just forgot to type [tex]\pm[/tex] in front of 12i

I had a chance to look at this on paper instead of while I was typing it. Here's what it looks like to me...

we have the equation [tex]y^2 + 22y + \sqrt{y} = -144[/tex]

because of the item [tex]\sqrt{y}[/tex] y can't be negative, meaning it has to be either an imaginary number or a positive number..

When I checked for possibilities of imaginaries, found the following:

*the left side will never be negative with positive numbers
*because of \sqrt{?i} we can't use imaginary numbers either
*already noted non-negative..

I didn't factor out the square root and then check imaginary numbers, so maybe solving polynomials with imaginary combinations or pwers is still possible, I don't know.
But I did go through the long process of working this out in a quartic formula, probably got lost somewhere in the math on paper but the only coefficients came out for me to be p = +-1 and q = +-1,2,4,8, which if I did it right came to the possible zero theorem check to +-1, +-1/2, +-1/4, +-1/8. I just checked the negatives and none work so I stoped and concluded that it wasn't solvable by a quartic function either....

Care to take a knab t it? I was trying all the usual ways of factoring polynomials. It's notable that every online tool I tried suggested the same - this function will never evaluate to -14 unless there is some way to utilize imaginary numerals...

This simple one proved to be so difficult for you, I presume that is why you find the difficult one so very very difficult.

Guest wrote:

Guest wrote:You seem to be an expert and know so much about algebra.....can you solve simple algebra problem for [tex]x[/tex]
[tex]x^{2} = -144[/tex]

Sure.. Like I said, I've been up late and algebra is not my strong card...

[tex]x^2 = -144[/tex]
[tex]\sqrt{x} = \sqrt{-144}[/tex]
Which is of course an error, can't square root a negative number..

But then if you wanted to go a little beyond algebra.. we can use imaginary numbers of course, but a calculator currently doesn't do this...

[tex]\sqrt{x}^2 = x = \sqrt{-144}[/tex]
[tex]x = \sqrt{-1*144} = \sqrt{-1*12*12} = \sqrt{-1}*\sqrt{12}^2[/tex]
[tex]x = 12i[/tex]

I'm tired now, getting off of here. Sorry for rushing things and distracting ppl... bye bye!

Guest wrote:You seem to be an expert and know so much about algebra.....can you solve simple algebra problem for [tex]x[/tex]
[tex]x^{2} = -144[/tex]

Sure.. Like I said, I've been up late and algebra is not my strong card...

[tex]x^2 = -144[/tex]
[tex]\sqrt{x} = \sqrt{-144}[/tex]
Which is of course an error, can't square root a negative number..

ugh... I'm sorry about the confusion there.

The solution I stated only works if the inverse exponents are the only two parts to the problem...

So the actual solution isn't that simple..

It's solvable on a plane. But it takes a while...

using the rules:

First step is to factor out the fraction...

[tex]y^{\frac12} = (-y^2 - 22y - 144)[/tex]
Then we simplify view of fraction and apply
[tex]\sqrt[2]{y}^1 = \sqrt[2]{y} = (-y^2 - 22y - 144)[/tex]
Simplify..
[tex]\sqrt[2]{y}^2 = y = (-y^2 - 22y - 144)^2[/tex]
[tex]y = (-y^4 - 22y^2 - 144^2)[/tex]
[tex]y = -y^4 - 22y^2 - 20,736[/tex]
[tex]20,736 = y^4 + 22y^2 + y[/tex]
Factor..
[tex]20,736 = y(y^3 + 22y + 1)[/tex]

This is as far as I can simplify it. Mathway suggests it solves to two rational numbers. Maybe someone else can take it from here... Every online tool I plugged this into otherwise suggested no solution.

You seem to be an expert and know so much about algebra.....can you solve simple algebra problem for [tex]x[/tex]
[tex]x^{2} = -144[/tex]

Guest wrote:How do you find that 22y = -144?

See my second post - I forgot to include 'y', and instead treated it as '0'... oops. I also wrote 144 instead of 114. That happens when I rush through algebra.. make mistakes

Sorry, in my earlier rush I did that wrong...

It's this:

[tex]y^2 + 22y + y^{\frac{1}{2}} = -114[/tex]
[tex]y^2 + \sqrt[2]{y} + 22y = -114[/tex]
[tex]y + 22y = -114[/tex]
[tex]23y = -114[/tex]
[tex]y = -114/23[/tex]
[tex]y = -4 \frac{22}{23}[/tex]

So in sum, the two parts [tex]x^2[/tex] and [tex]x^{\frac12}[/tex] cancel each-other out...

The same is true of anything similar... [tex]x^3 + x^{\frac13} = x^3 + \sqrt[3]{x} = x[/tex] and similarly [tex]x^{32} + x^{\frac{1}{32}} = x^{32} + \sqrt[32]{x} = x[/tex]...

It sounds like this is math homework, and so I'm happy to help - everyone should be able to master mathematics to a certain extent. From this example, it is too simple to simply cancel out exponents. So I'll provide another example so you now know how to do this with any exponential variable..

There's a lot of resources online to solve for exponentation. Algebra, though, is not my strong card...

How do you find that 22y = -144?

Guest wrote:Solve the equation:
[tex]y^2+ 22y + y^{\frac12}=-114[/tex]

The [tex]y^{\frac12}[/tex] is the same as [tex]\sqrt{y}[/tex]. So we can re-write the equation as (even though there are many more ways to solve this..)

[tex]y^2 + \sqrt{y} + 22y = -114; 22y = -114; y = -114/22; y = 5\frac{2}{11}[/tex]

Solve the equation:
[tex]y^2+ 22y + y^{\frac12}=-114[/tex]