1,2:
You have to find the roots of the quadratic equation, because
where are roots of the equation.
3:
It is known that the area of a right triangle is , because that is a half rectangle, where are the legs of the triangle.
I've done my homework this time and I think I have answers to some of my questions!
The thing is I don't know if I'm right.......
Factor the following polynomials.
1
x^2+x-6
I used brute force and got the answer (x+3)(x-2), I do not remember how to show working for this kind of question though.
2
6x^2+13x-5
Same deal here....Brute force, no idea how to show working.....
The area of a right triangle is 250 in^2. find the lengths of its legs if one leg is 5 inches longer than the other.(use the five step method)
I know the lengths: 20 in and 25 in, but as above I use brute force, so--yeah, I'm a little clueless as to how to show my work.
x^3+5x^2=6x
I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?
Another method to use when trying to factor a quadratic , is to multiply the coefficient of the term and the constant term (ac), then look for a pair of factors for this whose sum is the coefficient of the linear term (b). So for problem 1, ac=1(-6)=-6; so we look for two numbers whose product is -6 and whose sum is 1. Since ac=-6 is negative, one factor will be negative, and the other positive; so our problem is equivalent to finding factors of 6 whose difference is 1. Since the pairs of factors that multiply to 6 are 1 & 6 or 2 & 3, we see that 3-2=1, and so 3*(-2)=-6, and 3+(-2)=1. These two factors form the two x terms produced in FOIL that add to become the x term in the quadratic, so
.
Similarly for problem 2, ac=6*(-5)=-30, so we need a pair of numbers that multiply to -30 and add to 13; looking at pairs of factors of 30 and their differences:
30 & 1; difference=29
15 & 2; difference=13*
so we have 15*(-2)=-30, 15+(-2)=13,
so
.
--Kevin C.
ROFLMAO!
I forgot to add the answer I found to my second question, I see that I'm a little off with it, but I'll still post it:
2
6x^2+13x-5
(2x-5)(3x+1)
Same deal here....Brute force, no idea how to show working.....
Thank you guys, I think I understand how to work those questions!
WOOOO!!
One small thing:
What do I do here?
x^3+5x^2=6x
I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?