SOLUTION: The solutions to the equation 6x^2 + 10x = -2x^2 - 12x - 17 can be written in the form $x=\frac{P \pm \sqrt Q}{R}$, where $P$ and $R$ are relatively prime integers and $R > 0$.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The solutions to the equation 6x^2 + 10x = -2x^2 - 12x - 17 can be written in the form $x=\frac{P \pm \sqrt Q}{R}$, where $P$ and $R$ are relatively prime integers and $R > 0$.       Log On


   

Question 1209592: The solutions to the equation
6x^2 + 10x = -2x^2 - 12x - 17
can be written in the form $x=\frac{P \pm \sqrt Q}{R}$, where $P$ and $R$ are relatively prime integers and $R > 0$.
Found 4 solutions by greenestamps, Edwin McCravy, ikleyn, mccravyedwin:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


6x%5E2+%2B+10x+=+-2x%5E2+-+12x+-+17

Put the equation in standard form for a polynomial:

8x%5E2%2B22x%2B17=0

Plug the numbers into the quadratic formula:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

a=8, b=22, c=17

x+=+%28-22+%2B-+sqrt%28+22%5E2-4%2A8%2A17+%29%29%2F%282%2A8%29+

x+=+%28-22+%2B-+sqrt%28484-544%29%29%2F%2816%29+

x+=+%28-22+%2B-+i%2Asqrt%2860%29%29%2F%2816%29+

Simplify as necessary....

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The solutions to the equation
6x^2 + 10x = -2x^2 - 12x - 17
can be written in the form x=(P +/- sqrt(Q)/R, where P and R are relatively
prime integers and R > 0.
It would be better if would write your problems as above, and never use
the notation you used on here since it is not compatible with the HTML in which
this site was written.  You've been told this before, so I am starting to
wonder if perhaps you are not a human.  We know that many solutions answered
here are not done by humans.  Could some problems be posted by non-humans too?  

As the other tutor has shown, the solutions are non-real. They can be written
with Q being a negative integer.

x+=+%28-22+%2B-+sqrt%28-60%29%29%2F%2816%29+

However P=-22 and R=16 are not relatively prime, so even though you stated that
it would be possible, to the contrary, it turns out not to be possible.  

Edwin

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Edwin solved the problem practically to the end and derived the formula for the roots

    x%5B1%2C2%5D = %28-22+%2B-+sqrt%28-60%29%29%2F16.


But Edwin forgot to make the last step and to reduce this fraction.


So, I make this step on my own    

    x%5B1%2C2%5D = %28-11+%2B-+sqrt%28-15%29%29%2F8.


This time, the formula is presented in the form with P = -11 and R = 8, 
that are relatively prime integer numbers.


The numbers P and R are relatively prime integers, and the number R in the denominator is positive,
so all requirements are satisfied.

Solved.



Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!

She's right.  I was thinking if I simplified it there would then be an integer
multiplier in front of the square root and it would not be in the form 

x=%28P+%2B-+sqrt%28Q%29%29%2FR

but in the form x+=+%28P+%2B-+N%2Asqrt%28Q%29%29%2FR+

so I didn't bother to simplify the radicand because I didn't think N might
turn out to be 1 and I wouldn't have to write it.

Edwin