Why is 6√3 irrational or rational?
2 answers:
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Answer:
2
Step-by-step explanation:
So I'm going to use vieta's formula.
Let u and v the zeros of the given quadratic in ax^2+bx+c form.
By vieta's formula:
1) u+v=-b/a
2) uv=c/a
We are also given not by the formula but by this problem:
3) u+v=uv
If we plug 1) and 2) into 3) we get:
-b/a=c/a
Multiply both sides by a:
-b=c
Here we have:
a=3
b=-(3k-2)
c=-(k-6)
So we are solving
-b=c for k:
3k-2=-(k-6)
Distribute:
3k-2=-k+6
Add k on both sides:
4k-2=6
Add 2 on both side:
4k=8
Divide both sides by 4:
k=2
Let's check:
:
I'm going to solve for x using the quadratic formula:
Let's see if uv=u+v holds.
Keep in mind you are multiplying conjugates:
Let's see what u+v is now:
We have confirmed uv=u+v for k=2.