By how much does the larger root of the equation 2x^2+5x=12

[ghong14]

By how much does the larger root of the equation 2x^2+5x=12 exceed the smaller root?

a) 5/2

b) 10/3

c) 7/2

d) 14/3

e) 11/2

I know we can use the quadratic formula to solve for this problem. However, I took about 3 minutes trying to find the right combination to factor out the problem (x+6)(2x-2). Is there any suggestions that can help me find the 2 roots of this problem any faster?

[tim]

Let me see if I can derive a formula on the fly here:

Call the roots m and n. We know that the sum of the roots is -5/2 and the product is -12/2 (look up information on sums and products of roots if this isn't familiar to you), and we want the difference.

mn = -6
m+n = -2.5
we want m-n (assuming without loss of generality that m>n)

(m+n)^2 = (-2.5)^2
m^2 + 2mn + n^2 = 6.25
m^2 -12 + n^2 = 6.25
m^2 + n^2 = 18.25

Now we want m-n, so let's start with the squared version of it:

(m-n)^2 = m^2 - 2mn + n^2 = 18.25 + 12 = 30.25

Take the square root and you get the answer, 5.5. In general, here's how to get the difference (if you're interested in memorizing a super obscure formula):

(m-n)^2 = (m+n)^2 - 4mn

So to find m-n, we take the square root of (sum squared minus 4 times the product)

Personally, I'd just use factoring or, barring that, the quadratic formula to solve this one!

BTW your factored version of the expression is incorrect; you should be sure to check your work on factoring problems such as these.

[ghong14]

Yep you are absolutely correct the correct factor is (x+4)(2x-3). The other one was one of the combinations that did not work out during the course of my trial and error run.

[RonPurewal]

I know we can use the quadratic formula to solve for this problem.

Ok yeah. But, the question is this: What do you want to DO with the quadratic formula?

That might sound like a stupid question at first -- but, if your answer is "Find the two roots", then, nope.
If that's your answer, then you've completely forgotten the goal of the problem -- a problem that's distressingly common in any "academic" situation.

The GOAL of the problem is only to find the difference between the two roots. You don't have to find the roots themselves.

Consider the quadratic formula in this light. The two roots are
(stuff ± √(b^2 - 4ac)) / 2a
which can be distributed to
stuff/2a ± (√(b^2 - 4ac))/2a

The point here is that the "stuff/2a" part doesn't affect the difference at all, because it's the same in both terms. The part that matters is plus-or-minus blah blah blah.
Since that's the same, you know that the difference is just twice the plus/minus amount. (If you don't see why, just think about adding and subtracting 10 from something; the difference will be 20.)

So, the difference is
2 x (√(b^2 - 4ac))/2a
= (√(b^2 - 4ac))/a.

Plug in the numbers: (√(25 - 4*2*-12))/2
= (√(25 + 96))/2
= (√121)/2
= 11/2

--

Or yeah you could just factor the thing and get it over with.
(Interestingly, this is the only official problem I've ever seen that has used the factoring of a quadratic whose "x^2" coefficient is not just 1.)