To solve algebraic equations like t² - t³ = 12, rearrange terms to standard form, factor using algebraic identities (difference of squares and sum of cubes), and apply the quadratic formula to find all solutions, including real and complex roots.
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Solve for t in this nice Algebra equation | Math Olympiad MathematicsAdded:
In this video, let us solve for t given t raised to power 2 minus 2 raised to power 3 is equal to 12.
We are given t raised to power 2 minus t raised to power 3 is equal to 12.
Let us bring 12 to the left-hand side.
This will give us t raised to power 2 minus t raised to power 3 minus 12 is equal to 0.
Then we get t raised to power 2 minus t raised to power 3 then minus I'm going to split 12 here to give us 4 + 8.
Then if we open up this bracket to get t raised to power 2 minus t raised to power 3 minus 4 minus 8 is equal to 0.
We can also rewrite this as t raised to power 2 minus t raised to power 3 minus 4 here is 2 raised to power 2 then 8 is 2 raised to power 3.
is equal to 0.
I'm going to rearrange this by order of power so that we have t raised to power 2 minus 2 raised to power 2 then minus t raised to power 3 minus 2 raised to power 3 is equals to 0.
This would then imply t raised to power 2 minus 2 raised to power 2 minus t raised to power 3 plus 2 raised to power 3 is equals to 0.
From here we have addition of two cubes.
But here we have difference of two cubes.
By identity given a raised to power 2 minus b raised to power 2 we can factorize this to give us a minus b times a plus b.
So this here becomes t minus 2 times t plus 2.
Then minus t raised to power 3 plus 2 raised to power 3 is equals to 0.
As mentioned earlier, this is addition of two cubes.
By identity given a raised to power 3 plus b raised to power 3 we can factorize this as a plus b into brackets a squared minus ab plus b squared.
So using these two factorize this, we're going to have t raised to power 3 plus 2 raised to power 3 is equals to t plus 2 into bracket t squared minus 2t.
Then + 2 squared.
So, we're going to put this in place of this here to give us t - 2 * t + 2 then minus t + 2 into bracket t squared - 2t. This will be a + 4.
Then equal to 0.
We have t + 2 here, t + 2 here. We can factorize that to give us t + 2 into bracket t - 2 then minus all of this t squared minus 2t + 4 then equal to 0.
We get t + 2 into bracket t - 2 - t squared + 2t - 4 is equal to 0.
t + 2 into brackets I have t here I have 2t here giving us 3t then minus 2 minus 4 that gives us minus 6 and then minus t squared is equal to 0 which will then imply that either t + 2 is equals to 0 or 3t - 6 - t squared is equals to 0.
From the first equation, which is a linear equation, we get t is equals to -2.
We'll call this t1.
And then the second equation, which appear to be a quadratic equation, we can rewrite as -t squared + 3t -6 is equals to 0.
We divide through by -1, this will give us t squared -3t +6 is equals to 0.
Now, we solve this quadratic equation for two more values of t.
To solve for t here, let's first compare with the general quadratic equation ax squared + bx + c equals to 0.
So, that we get the values of a, b, and c.
a is 1, b is -3.
And c is 6.
Then, we go ahead and solve for t using the general quadratic formula, which is -b + - square root of b squared -4ac then divided by 2a.
We have our values for a, b, and c. So, let's plug into this formula.
This will give us t is equals to -b which is -3 + square root of minus three squared minus four times one times six then divided by two A which is two times one.
This will give us T is equal to positive three plus or minus square root of negative three squared here will give us nine minus four times one is four four times six is 24.
Square root of that then divided by two times one is two.
So this gives us T is equal to three plus or minus square root of nine minus 24 will give us negative 15.
Then divided by two.
So we have T is equal to three plus or minus square root of this will be 15 times negative one divided by two.
If we separate the radical we get T is equal to three plus or minus square root of 15 and then times square root of negative one then divided by two.
This will give us T is equal to three plus or minus This will be I square root of minus one is I then root 15.
Divided by two.
Let us separate this division here. This will give us T is equal to 3 over 2 plus or minus I root 15 divided by 2.
And then we split this into two different values of T.
Recall that earlier we got a value for T as T1 which gave us -2.
And now here we are going to have T2 equal to 3 over 2 plus I root 15 divided by 2 and T3 equal to 3 over 2 minus I root 15 divided by 2.
These are all possible solutions from this problem. We have two complex roots and only one real solution.
The next thing we are going to do now will be to do a quick check with the real solution.
The problem given was T raised to power 2 minus T raised to power 3 is equal to 12.
And now we got T1 which is the only real solution to be -2.
So, I'm going to put in place of T in this equation -2.
This would then imply -2 raised to power 2 minus -2 raised to power 3 to give us 12.
-2 raised to power 2 give us 4.
- -2 raised to power 3 will give us -8.
That will give us 12.
This will give us 4 + 8 is equals to 12.
4 + 8 is 12, so we have 12 is equal to 12.
And since the left-hand side balances the right-hand side, that confirms that the solution we got for T here is absolutely correct.
Thanks for watching. Please like and share. And also remember to subscribe to my channel.
And I'll see you in my next video.
Bye.
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