How do you solve a quadratic equation using the completing the square method?

  • Thread starter Betamax
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In summary, guys, I am trying to isolate the time variable in this standard gravity equation (without time appearing on both sides) and cannot do it. Could one of you please help and explain how. the seperating of the two time bits on the right is what is stumping me.
  • #1
Betamax
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Guys,

I have spent too long on this my head hurts!

I am trying to isolate the time variable in this standard gravity equation (without time appearing on both sides) and cannot do it. Could one of you please help and explain how. the seperating of the two time bits on the right is what is stumping me.

S = ut + 1/2at^2


Thanks in advance
 
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  • #2
A falling object is an object that you drop from some height. its initial velocity is zero (xi = 0).
There are simple derived equations that allow you to calculate the velocity and distance traveled, as well as the time taken to achieve a given velocity or distance.
 
  • #3
willsmith123 said:
A falling object is an object that you drop from some height. its initial velocity is zero (xi = 0).
There are simple derived equations that allow you to calculate the velocity and distance traveled, as well as the time taken to achieve a given velocity or distance.

Yes I know. If I wanted to calculate the time to chuck a lead weight upwards at 3m per second, I want the equation that says how long to get to its apex. To do that, I need to isolate the time component. Being shown how would be a bonus.
 
  • #4
Betamax said:
I am trying to isolate the time variable

This is a quadratic equation for t. Have you heard of the "quadratic formula"?

Generically, if [itex]ax^2 + bx + c = 0[/itex], then

[tex]x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
 
  • #5
jtbell said:
This is a quadratic equation for t. Have you heard of the "quadratic formula"?

Generically, if [itex]ax^2 + bx + c = 0[/itex], then

[tex]x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

Yes I have. I left school 37 years ago!

Can you tell me the time equation please? I don't quite get the -b part or the + or - part in your equation btw.
 
  • #6
jtbell said:
This is a quadratic equation for t. Have you heard of the "quadratic formula"?

Generically, if [itex]ax^2 + bx + c = 0[/itex], then

[tex]x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

There are also 3 addition parts to this equation, so I have no idea how to isolate the time component.
 
  • #7
Your original equation, rewritten is:

(1/2 a) t^2 + (u) t - S = 0

The general from of the quadratic equation is

A t^2 + B t + C = 0

So, your equations is an example of the quadratic equation with:

A = 1/2 a
B = u
C = -s

This should give you the info you need to solve your equation. It might be helpful to review how the quadratic equation was solved, by "completing the square".

We start with

A t^2 + B t + C = 0

Consider A*(t+B/2A)^2. We can expand this to get

A t^2 + A 2 t B/2A + A B^2/4A^2 = At^2 + B t + B^2 / 4A

So we can write A t^2 + B t + C = A(t+B/2A)^2 -B^2 / 4A + C

A (t+B/2A)^2 = B^2 / 4A - C= (B^2 - 4AC )/ 4A

By taking the square root and dividing, we can solve for t+B/2A, and then for t. This is known as "completing the square" - but people generally don't do this explicitly, they simply use the quadratic formula when they see a quadratic equation.
 

FAQ: How do you solve a quadratic equation using the completing the square method?

What is the gravity equation conversion?

The gravity equation conversion is a mathematical formula used to calculate the force of gravity between two objects. It takes into account the mass and distance between the objects to determine the strength of the gravitational pull.

How is the gravity equation conversion used?

The gravity equation conversion is used in many fields of science, such as astrophysics and engineering, to calculate the force of gravity between objects. It is also used in everyday applications, such as predicting the motion of objects in orbit or designing structures that can withstand gravitational forces.

What are the units of measurement used in the gravity equation conversion?

The units of measurement used in the gravity equation conversion are typically the metric units of kilograms for mass and meters for distance. However, other units may be used depending on the specific application.

How accurate is the gravity equation conversion?

The gravity equation conversion is considered to be highly accurate when used correctly. However, it may not account for all factors that could affect the force of gravity, such as the curvature of space-time in extreme cases.

Are there any limitations to the gravity equation conversion?

While the gravity equation conversion is a useful tool for calculating the force of gravity, it has limitations. It assumes that the objects are point masses and does not take into account the effects of other forces, such as air resistance. Additionally, it may not be accurate for extremely large or small objects or in situations with high levels of gravity, such as near black holes.

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