Quadratic Equation help


by RStars
Tags: equation, quadratic
RStars
RStars is offline
#1
Apr28-12, 05:24 PM
P: 7
Hey,

I was going through some mechanics notes and came across this quadratic equation to solve for x. In my notes it is supposed to equal 3 however I do not get that result. I am not sure if I am simplifying it wrong or what. I am ending up with 2.12 for the positive value. Please let me know if you get 3 or 2.12 so that I know if the error is in my calculations or in my notes.



Thanks in advanced.
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DonAntonio
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#2
Apr28-12, 05:39 PM
P: 606
Quote Quote by RStars View Post
Hey,

I was going through some mechanics notes and came across this quadratic equation to solve for x. In my notes it is supposed to equal 3 however I do not get that result. I am not sure if I am simplifying it wrong or what. I am ending up with 2.12 for the positive value. Please let me know if you get 3 or 2.12 so that I know if the error is in my calculations or in my notes.



Thanks in advanced.


It must be some mistake in your notes: if you put [itex]x=3[/itex] in the given eq., one gets that the LHS is not an integer

whereas the RHS is...

DonAntonio
Borek
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#3
Apr29-12, 02:18 AM
Admin
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Neither 3 nor 2.12 are roots of the equation as you wrote it.

Integrand
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#4
Apr29-12, 02:36 AM
P: 16

Quadratic Equation help


3 is correct if the LHS is amended to contain 144x2 rather than 144 + x2.
evilbrent
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#5
Apr29-12, 06:29 PM
P: 4
um... I haven't done this in a while so forgive me for being simple.... but that's not an equation.

if you put x=10 you end up with the equation 122=900, which isn't true.

There's been some kind of mistake.
DonAntonio
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#6
Apr29-12, 09:22 PM
P: 606
Quote Quote by evilbrent View Post
um... I haven't done this in a while so forgive me for being simple.... but that's not an equation.

if you put x=10 you end up with the equation 122=900, which isn't true.

There's been some kind of mistake.


No, that is too an equation. To solve it means to find out the numerical values of x that when substituted in the equation give

a true equality. What you've shown above is that the numerical value x = 10 is not (one of the) a solution(s) of the equation.

DonAntonio
evilbrent
evilbrent is offline
#7
Apr29-12, 09:33 PM
P: 4
oh, ok, yes. Sorry engineering maths was a decade ago for me. It's amazing how quickly the knowledge vanishes.

I reduced the original equation down to 0=-71.5x^2+900x-2628 and got 0=(x-4.604)*(x-7.984)


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