Solving a Quadratic Expression: How to Solve 2x^2+9x = -5

AI Thread Summary
To solve the quadratic equation 2x^2 + 9x = -5, it is essential to first rearrange it into standard form, ax^2 + bx + c = 0. This allows the use of the quadratic formula to find the roots. The original poster struggled with the problem, feeling confused and unable to factor or solve it initially. After receiving guidance on using the quadratic formula, they successfully solved the equation and expressed relief at resolving their confusion. The discussion highlights the importance of understanding the standard form and the quadratic formula in solving quadratic equations.
Trent21
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2x^2+9x = -5

Solved, you can banish this thread now mods.




2x^2+9x = -5
With the given answers, nothing I can come up with matches it. I can't solve it like a normal quadratic, as it gives me no answer that matches those. And I can't factor it. Nothing that adds up to 9, that when you also multiply them, you get -5.

Divide off the 2...?
Divide off the 9.
Now I've got X + X or X^2 on the left side, and that's already not a answer match, can't solve it for X.
Cant factor it.

Square rooting it just gives The suare root of 5 divided by something, that's not a answer. So I am done, not including I feel pretty dumb not being able to solve this basic math but I am doing Thermodynamics at the moment. I can't remember what to do with this...

Only thing I heard of was to get it in standard form or something. Nothing I learned. Its this little problem that I can't figure out...

Could someone explain how to solve this, according to the answers given, and why it works? I asked others, even other friends but even they don't know, even asked a Cop at the store the other day and he didn't know. And I can't find any questions exactly like this online, is this question that tedious? Its not even difficult I bet.

Thanks
 
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The "standard form" you spoke of is ax^2+bx+c=0 (you just get all the terms on to one side)

The reason you would want to do this is so you can use the quadratic equation. (If you don't remember the quadratic equation, it should be very easy to look up.)

As to why the quadratic equation works? It is just a generalized solution for an equation of this form (I don't want to derive it for you, but, if you're curious, that should also be easy to look up)



P.S.
In the future, the math-homework section of these forums will probably give you more help on purely math problems.
 
Nathanael said:
The "standard form" you spoke of is ax^2+bx+c=0 (you just get all the terms on to one side)

The reason you would want to do this is so you can use the quadratic equation. (If you don't remember the quadratic equation, it should be very easy to look up.)

As to why the quadratic equation works? It is just a generalized solution for an equation of this form (I don't want to derive it for you, but, if you're curious, that should also be easy to look up)



P.S.
In the future, the math-homework section of these forums will probably give you more help on purely math problems.

Ooooooooooh hahaha, I feel pretty dumb I forgot all about the quadratic formula. I just plugged all this right in after putting it in standard form and that did the trick.

Im all good now. I am going to edit my post away so I don't embarass myself even further, thanks Nathan. : D
 
Trent21 said:
Ooooooooooh hahaha, I feel pretty dumb I forgot all about the quadratic formula. I just plugged all this right in after putting it in standard form and that did the trick.

Im all good now. I am going to edit my post away so I don't embarass myself even further, thanks Nathan. : D

Welcome to the PF.

It is against the PF Rules to edit your OP after you have solved the problem. Please leave it as-is in the future.
 
I restored the post, helpers please remember to first quote the OP before responding, thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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