Question

Solve the equation by completing the square , t^2-10t-24=0

Answer 1

Well, do it. What's first?

Answer 2

The question is wrong it will be +24 :/

Answer 3

Not sure why you would know that, but it makes no difference for the methodology. Let's see it!

Answer 4

that is what it says so

Answer 5

i have four others like tht so im lost

Answer 6

It remains of no consequence. Please demonstrate the method. What's first?

\(t^{2}-10t-24=0\)

Needs to look like this:

\((t-Something)^{2} = Something\;Else\)

Answer 7

u have to set all up on the left side it says thats it i have no clue it for squares

Answer 8

You need only this principle \((a+b)^{2} = a^{2} + 2ab + b^{2}\)

Answer 9

how u do this then

Answer 10

expanding on that principle ...say a is t
\[(t+b)^{2} = t^{2} +2bt +b^{2}\]
now ignore the constant (-24) for now
\[t^{2} -10t + ? = t^{2} +2bt +b^{2}\]
what does "b" need to be to make it fit (turn it into a perfect square)
-10 = 2b --> b = -5
---> b^2 = 25
now you have
\[t^{2} -10t +25 = (t-5)^{2}\]
thats completing the square
ok to keep everything balanced, we need to subtract 25 since we added it to make the perfect square
\[\rightarrow t^{2}-10t = (t^{2} -10t +25)-25 = (t-5)^{2} -25\]
now add on the constant we ignored earlier
\[\rightarrow (t-5)^{2} -25 -24 = 0\]
now solve for t
\[(t-5)^{2} = 49\]

Answer 11

Take a good, hard look at this:

\((t-a)^{2} = t^{2} - 2at + a^2\)

Now, take a good hard look at this:

\(t^{2}-10t-24=0\)