1. Apr 28, 2012

### RStars

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.

http://img707.imageshack.us/img707/338/codecogseqno.gif [Broken]

Last edited by a moderator: May 5, 2017
2. Apr 28, 2012

### DonAntonio

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

whereas the RHS is...

DonAntonio

Last edited by a moderator: May 5, 2017
3. Apr 29, 2012

### Staff: Mentor

Neither 3 nor 2.12 are roots of the equation as you wrote it.

4. Apr 29, 2012

### Integrand

3 is correct if the LHS is amended to contain 144x2 rather than 144 + x2.

5. Apr 29, 2012

### evilbrent

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.

6. Apr 29, 2012

### DonAntonio

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

7. Apr 29, 2012

### evilbrent

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)