Respuesta :
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:
[tex]3x^2-(3k-2)x-(k-6) \text{ with }k=2[/tex]:
[tex]3x^2-(3\cdot 2-2)x-(2-6)[/tex]
[tex]3x^2-4x+4[/tex]
I'm going to solve [tex]3x^2-4x+4=0[/tex] for x using the quadratic formula:
[tex]\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}[/tex]
[tex]\frac{4\pm \sqrt{16-16(3)}}{6}[/tex]
[tex]\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}[/tex]
[tex]\frac{4\pm 4\sqrt{-2}}{6}[/tex]
[tex]\frac{2\pm 2\sqrt{-2}}{3}[/tex]
[tex]\frac{2\pm 2i\sqrt{2}}{3}[/tex]
Let's see if uv=u+v holds.
[tex]uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}[/tex]
Keep in mind you are multiplying conjugates:
[tex]uv=\frac{1}{9}(4-4i^2(2))[/tex]
[tex]uv=\frac{1}{9}(4+4(2))[/tex]
[tex]uv=\frac{12}{9}=\frac{4}{3}[/tex]
Let's see what u+v is now:
[tex]u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}[/tex]
[tex]u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}[/tex]
We have confirmed uv=u+v for k=2.
