so....here it is......
"Find the rational numbers a,b so that one of the two solutions of the equation x^2+ax+b=0 is the number 2-sqr5.Then, find the other solution of this equation"
I have been trying to solve this equation for hours!!!!!!!!!!!!!!!!!!!!!!!!!... help me!!!
2007-01-28 06:10:36 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2 + ax + b = 0
Given one of the solutions is 2 - sqrt(5).
If we're given that we have rational coefficients, we have to be aware that irrational roots come in conjugate pairs. The other solution should be 2 + sqrt(5).
Since 2 - sqrt(5) is a root, it follows that [x - (2 - sqrt(5))] is a factor.
Similarly, if 2 + sqrt(5) is a root, [x - (2 + sqrt(5)] is a factor. So we have
[x - (2 - sqrt(5))] [x - (2 + sqrt(5))]
Expand this,
x^2 - x[2 + sqrt(5)] - x[2 - sqrt(5)] + [2 - sqrt(5)][2 + sqrt(5)]
x^2 - x[2 + sqrt(5) + 2 - sqrt(5)] + [4 - 5]
x^2 - x[4] - 1
x^2 - 4x - 1
As you can see, a = -4 and b = -1.
2007-01-28 06:22:32 · answer #1 · answered by Puggy 7 · 1⤊ 0⤋
Uh, wat grade r u in?
I learned this in 5th grade...but i forgot...
i'm a 6th grader now.
we're gonna learn it again in Feb.
I'm sorry I can't help u.
I think u need to- no, no!
I'm sorry, i keep forgetting wat 2 do!
2007-01-28 06:17:37 · answer #2 · answered by Anonymous · 0⤊ 2⤋