Solve y=x^2 & y=x-1/4 - Need Help

  • Thread starter Thread starter Kandy
  • Start date Start date
  • Tags Tags
    System
AI Thread Summary
To solve the equations y=x^2 and y=x-1/4, first set them equal to each other, resulting in the equation x^2 = x - 1/4. Rearranging gives a quadratic equation, which can be solved using the quadratic formula. Factoring fractions is similar to integers, requiring combinations that multiply to 1/4. An alternative approach is to graph the equations to find their intersection points. Understanding these methods can simplify solving quadratic equations involving fractions.
Kandy
Messages
27
Reaction score
0
y=x^2
y=x-1/4

I really need help with this question. i don't know how to do these types of questions with fractions. i can't think, i need to sleep :zzz:
 
Physics news on Phys.org
Edit: I jumped a step. Both equations are equal to y, so x^2=x-\frac{1}{4}

Get everything onto the same side of the equation. You have a quadratic equation.

Factoring fractions really isn't that much different than integers. You figure out the combinations that will equal 1/4 when multiplied together. For example:

\frac{1}{4}* 1 = \frac{1}{4}
\frac{1}{2}*\frac{1}{2} = \frac{1}{4}

and so on. Add the combinations together and hopefully one of them will equal your middle coefficient (1 in this case).
 
Last edited:
BobG said:
Get everything onto the same side of the equation. You have a quadratic equation.
Then, you can just use the quadratic formula to solve for the roots.
 
Kandy said:
y=x^2
y=x-1/4

I really need help with this question. i don't know how to do these types of questions with fractions. i can't think, i need to sleep :zzz:

Note that y=x-1/4 can be rewritten as 4y=4x-1. (No fractions here!)

As BobG suggests, since y=x^2 replace y in (4y=4x-1) by x^2:
4(x^2)=4x-1.

Then, as EnumaElish suggests, use the quadratic formula [after writing it in standard form].


An alternate method is to plot the two curves (one is a parabola and one is a line) then locate the intersection points.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top