SOLUTION: Finding finding the roots of a quadratic equation with leading coefficient greater than 1 Solve for u 3u^2-5u=2

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Question 693410: Finding finding the roots of a quadratic equation with leading coefficient greater than 1
Solve for u
3u^2-5u=2

Answer by pmatei(79) About Me  (Show Source):
You can put this solution on YOUR website!
For quadratics you need everything on the left side of the equal sign:
3u%5E2-5u-2=0
There are 2 ways of solving quadratics.
A. You try to find to numbers that their product is equal to the product of the u^2 coefficient and the number without letter (in this case 3%28-2%29=-6), and their sum is equal to the u coefficient (in this case -5). The numbers that multiplied give -6 and added give -5 are : -6 and %2B1.
Instead of -5u in the original equation write the two numbers multiplied by u:
3u%5E2+-6u%2Bu-2=0
Group the four terms in 2 pairs: first two terms and last two terms.
%283u%5E2-6u%29%2B%28u-2%29=0
See if you can factor out anything from those parentheses (you always have at least the letter from the first parenthesis):
3u%28u-2%29%2B1%28u-2%29=0
If you done everything right the two parentheses are the same at this point. This repeated parenthesis is one of the 2 parentheses you want to solve the quadratic. The second one is put together by what you factored out:
%28u-2%29%283u%2B1%29=0
For the quadratic solutions take each parenthesis and set it equal to zero.
First solution:
u-2=0
u=2
Second solution:
3u%2B1=0
u=-1%2F3
This method does not always work.
B. The second method is to know the quadratic formula, and works every time:
If ax%5E2%2Bbx%2Bc=0 then the solutions are given by:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In your case 3u%5E2-5u-2=0, a=3, b=-5, c=-2.
u=%285+%2B-+sqrt%28+25-4%2A%283%29%2A%28-2%29+%29%29%2F6+
u=%285+%2B-+sqrt%28+25%2B24%29%29%2F6+
u=%285+%2B-+sqrt%28+49%29%29%2F6+
u=%285+%2B-+7%29%2F6+
First solution:
u+=+%285%2B7%29%2F6+=+12%2F6=2
Second solution:
u=%285-7%29%2F6=-2%2F6=-1%2F3