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Finding a Absolute Value Quadratic - Math Team


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I got another math team packet, this is in the advanced math section, i.e., Trig and Calculus, but I still want to know how to do it.

So lets see, I need to solve the Quadratic for all values of X. The equation is |x| = x2+x-3=0.

So I see that is in Quadratic form. So I get the coefficients of the variables, plus the -3 at the end(ax2+bx+c=0)

So a = 1, b = 1, and c = -3. So I plug that into the Quadratic Formula.

This gets me to: [math]/frac{-1 \pm \sqrt13 }{2} = |x|[/math]

 

So |x| = -1 +- sqr(13)/2

 

The Math Drawer isn't really working...

 

Don't know were to go now

Edited by Lightmeow
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I got another math team packet

 

 

Shouldn't this be in the homework section?

 

Here is a hint.

 

Go back to the beginning and ask yourself

 

What makes you think b=1?

 

A modulus often means you have two equations to solve.

Edited by studiot
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I got another math team packet, this is in the advanced math section, i.e., Trig and Calculus, but I still want to know how to do it.

 

So lets see, I need to solve the Quadratic for all values of X. The equation is |x| = x2+x-3=0.

 

So I see that is in Quadratic form. So I get the coefficients of the variables, plus the -3 at the end(ax2+bx+c=0)

 

So a = 1, b = 1, and c = -3. So I plug that into the Quadratic Formula.

This gets me to: [math]/frac{-1 \pm \sqrt13 }{2} = |x|[/math]

 

So |x| = -1 +- sqr(13)/2

 

The Math Drawer isn't really working...

 

Don't know were to go now

 

Sorry - I am lost...

 

You write

[latex]|x|=x^2+x-3=0[/latex]

you have a set of equations (there are two equals sign) which ignoring the central section - you must be able to do this - gives you

[latex]|x|=0 [/latex]

 

Now if the [latex]=0[/latex] is ignored you end up with the much more sensible

[latex]|x|=x^2+x-3[/latex]

And to solve this you need to rearrange to a format congruent with

[latex]ax^2+bx+c=0[/latex]

As Studiot hints - you have done this incorrectly. I believe merely by adding [latex]=0[/latex] rather than properly rearranging. Do it by the book and as per Studiots message you can get two equations (think what absolute value entails and which of the coefficients of the quadratic will be affected)

 

and Yes - the equation renderer - LaTex is screwy

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If |x|= x^2+ x- 3, then either x>= 0 or x< 0. if x>= 0 then |x|= x= x^2+ x- 3 so that x^2= 3. That has two solutions but only one is non-negative.

 

If x< 0, the |x|= -x= x^2+ x- 3 so that x^2+ 2x+ 3= (x+ 2))(x+ 1)= 0. That has two negative solutions.

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Sorry I haven't gotten back in a while on this topic. I found the answer, it was -3 and the square root of 3. I was just getting hung up on the fact that the question was |x|=x2+x-3, so it wasn't in quadratic form, which is ax2+by+c=0. Oh, and for studiot, I know that b is 1 because x is really 1x, and one is the coefficient of x.

 

Thanks for the help

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I found the answer, it was -3 and the square root of 3. I was just getting hung up on the fact that the question was |x|=x2+x-3, so it wasn't in quadratic form, which is ax2+by+c=0. Oh, and for studiot, I know that b is 1 because x is really 1x, and one is the coefficient of x.

 

 

Well I'm pleased to receive a response, even if it was not a correct one.

 

If |x|=x2+x-3

 

Then rearranging

 

x2+x-|x|-3 = 0

 

x2+(x-|x|)-3 = 0

 

x2+(x-x)-3 = 0 : x>0

 

x2+(x-(-x))-3 = 0 : x<0

 

02+(0-0)-3 = 0 : x=0 which implies x cannot be zero

 

Thus there are, as I said, two equations to solve.

 

One does indeed lead to x = +/- sqroot(3), but subject to the the condition x>0 so we must discard the negative root and only accept the positive. You can check this by substitution into the original expression.

 

I think Halls made transposition error in his second equation, somehow changing the sign of the 3. (post#4)

 

thus the second equation factorises as (x+3) (x-1) = 0 with the condition that x<0

 

This time we must discard the positve root and accept the negative x=-3

again we can check this by substitution.

 

So this quadratic does indeed lead to two solutions as expected.

 

 

But then you made a mistake in your first presentation (post#1) since you stated that |x| = x2+x-3=0.

 

Which implies that |x| = 0.

Which implies that x = 0.

 

Again I ask, does b ever = 1?

 

go well

 

Edited by studiot
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OK, b does never equal one. It equals 2 or 0. I must of gotten lucky when I did this problem.

 

Thanks for the help peoples. I can't wait when I am a senior and in AP calculus and am sad that I had this question.smile.png

 

Joshua

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Doing algebra with moduli is more tricky than ordinary algebra so I'm suprised to find it on the High School syllabus - it didn't used to be.

 

Anyway I thought you hung in there rather well.

 

Keep it up for the future and you will do well.

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  • 3 years later...
On ‎12‎/‎7‎/‎2013 at 11:59 AM, HallsofIvy said:

If |x|= x^2+ x- 3, then either x>= 0 or x< 0. if x>= 0 then |x|= x= x^2+ x- 3 so that x^2= 3. That has two solutions but only one is non-negative.

 

If x< 0, the |x|= -x= x^2+ x- 3 so that x^2+ 2x+ 3= (x+ 2))(x+ 1)= 0. That has two negative solutions.

It has been pointed out to me that -x= x^3+ x- 3 becomes x^2+ 3x- 3= 0, not "+3" as I had. That was a typo.  And neither of those factors as (x+2)(x+ 1)!  The solution to that quadratic equation, using the quadratic formula, is x= (-3+/- sqrt{9+ 12}/2= (-3+/- sqrt{21}/2.

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