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1. Please help me with this, I've tried for way too long on this, dont ask me to show my working because I've already wasted 1 hour on this and don't want to waste another minute now. Thanks.
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2. ignore me im an idiot

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3. (Original post by Jas1947)
Please help me with this, I've tried for way too long on this, dont ask me to show my working because I've already wasted 1 hour on this and don't want to waste another minute now. Thanks.
If you let , the equation becomes which you can solve.

Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
4. (Original post by RDKGames)
If you let , the equation becomes which you can solve.

Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
Oh my god i remember now! I cant believe i wasted so much time, thank you!
5. (Original post by RDKGames)
If you let , the equation becomes which you can solve.

Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
I was trying to figure out where I went wrong, I thought product meant multiplication so I was multiplying the roots xD

Edit: read the question again and it says sum and not product. ok then im seeing things

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6. (Original post by HFancy1997)
I was trying to figure out where I went wrong, I thought product meant multiplication so I was multiplying the roots xD

Edit: read the question again and it says sum and not product. ok then im seeing things

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Take a break from maths at 1am, mate.
7. For a more elegant method, gives so that where are the roots. Then taking gives
8. (Original post by Zacken)
For a more elegant method, gives so that where are the roots. Then taking gives
So that's how you do it. Back then at midnight I started thinking how the heck I could apply the alpha beta stuff to this and you've shown it, so thanks.
9. (Original post by RDKGames)
So that's how you do it. Back then at midnight I started thinking how the heck I could apply the alpha beta stuff to this and you've shown it, so thanks.
No problem. Motivating thing here is that product of roots of a polynomial is easy and logarithms turn products into sums.

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