Michael W. answered • 03/26/15
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Jose,
I read the question as:
9u - 7*(3u) + 12 = 0
The trick here is to make this look like a quadratic equation, which we can then hopefully solve.
- Can we make 9u look more like 3u? Well, if they had the same base, then they'd be related to each other. 9 is 3 squared.
- If 9 is 3 squared, then 9u is (32)u. Based on what you know about exponents, isn't that the same as 32u?
- What you have now is a quadratic in hiding. If you think about 3u as "x," then the first term is x2 (3u * 3u). The second term is -7x. And the third term is 12.
That's x2 - 7x + 12 = 0. It's a quadratic, where we have temporarily replaced 3u by x.
If we can factor it, then we'll end up with two equations: x = something, and x = something else.
But we're not quite done. The original problem was to solve for u, and we swapped out u. We know x. If 3u = x, then can we use our values of x to find the corresponding values for u?
Hope this helps!
Michael W.
That's the basic idea, yes. You treat 3u as "x" and then write it like a quadratic formula:
x2 - 7x + 12 = 0
Solve that for x as you would solve any other quadratic...you can use the quadratic formula, or see if factoring works.
Either way, you'll figure out what x has to be equal to. Let's say one of the solutions is x = 4. Then, you're still not done, because the question was to find u. You've found x. You're not quite there...but we were treating 3u as x. So,
we can work backwards and get the values for u.
If 3u equals x, and x equals 4, then:
3u = 4
That's solvable using a log?
You can do the same thing for the other value of x...and you have your two solutions for u.
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03/26/15
Michael W.
Oh, and just to be clear: you write it like a quadratic equation, not formula. :) The quadratic formula is for solving quadratic equations. Technicality, but I didn't want to encourage the wrong vocabulary...
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03/26/15
Jose G.
03/26/15