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

[Betamax]

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

 

[willsmith123]

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.

 

[Betamax]

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.

 

[jtbell]

I am trying to isolate the time variable

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

Generically, if ax^2 + bx + c = 0, then

x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

 

[Betamax]

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

Generically, if ax^2 + bx + c = 0, then

x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

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.

 

[Betamax]

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

Generically, if ax^2 + bx + c = 0, then

x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

There are also 3 addition parts to this equation, so I have no idea how to isolate the time component.

 

[pervect]

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.