Solve for x when given a trinomial?

2008-12-25 1:18 pm
How would one go about solving problems like this:

x^2 + 2x - 24 = 0

回答 (10)

2008-12-25 1:25 pm
✔ 最佳答案
The usual way to solve is to factor, and then equate each factor to 0 to find the solution(s).
x² + 2x - 24 = 0
(x + 6)(x - 4) = 0
(x + 6) = 0
x = -6
(x - 4) = 0
x = 4
x = {-6, 4}
2016-11-06 10:35 pm
Divide the two aspects by utilising -2 to make the full concern greater practicable. 10x^2 + 11x - 6 = 0 discover the aspects: (5x - 2)(2x + 3) = 0 Set each and every of those aspects to equivalent 0: (i) If 5x - 2 = 0 5x = 2 x = 2/5 (ii) If 2x + 3 = 0 2x = -3 x = -3/2 answer: x = 2/5, -3/2 –––––––––––––––––
2016-09-28 1:25 pm
Trinomial Solver
2008-12-25 11:23 pm
x^2 + 2x - 24 = 0
you have to first simplify it into two different equations
(x+6)(x-4)= x^2 + 2x - 24
then you solve for an x which would make the answer equal to zero
all you have to do is solve each equation for the number that will make the result equal to zero
1st | -6 + 6 = 0
2nd | 4 - 3 = 0
therefore
x = -6 and x = 4
2008-12-25 6:57 pm
(x+6) * (x-4) = 0

So x would be equal to -6 and 4.
2008-12-25 2:21 pm
x^2 + 2x - 24 = 0
x^2 + 6x - 4x - 24 = 0
(x^2 + 6x) - (4x + 24) = 0
x(x + 6) - 4(x + 6) = 0
(x + 6)(x - 4) = 0

x + 6 = 0
x = -6

x - 4 = 0
x = 4

∴ x = -6 , 4
2008-12-25 1:39 pm
If one can factor the polynomial in a reasonable amount of time, then that would be the first approach. If your trinomial happens to be quadratic, regardless of whether or not it factors, you can always use the quadratic formula to solve the equation.

For instance: Solve x^4+7x^2+16=0.

Note that x^4+7x^2+16=(x^4+8x^2+16)-x^2 = (x^2+4)^2-x^2 which is a difference of squares. Thus we have

(x^2-x+4)(x^2+x+4)=0

Neither of these quadratic factors factor, so we shall apply the quadratic formula.

We then obtain x=( pm 1 pm (i *sqrt (15)))/2 as the 4 solutions to the equation.

In general if your trinomial is not quadratic (and even when it is), check for a factorization. Not all polynomials factor (in fact the probability of choosing a reducible polynomial at random from the set of all polynomials is 0) but textbook problems are typically carefully chosen.
2008-12-25 1:34 pm
factor it so you get
(x-4)(x+6)=0
so either x-4=0 or x+6=0
so x=4 or -6
2008-12-25 1:24 pm
( x + 6 ) ( x - 4 ) = 0
x = - 6 , x = 4
2008-12-25 1:21 pm
you use factoring, it would factor to
(x+6)(x-4)=0
then you solve for x
so your answer will be
x=6 x=4


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