Question

If \[\alpha \neq \beta\] and \[3\alpha ^{2}-h \alpha-b=0\]\[3\beta ^{2}-h \beta-b=0\], then \[\alpha+ \beta\]=?
A. -\frac{ b }{ 3 }
B. \frac{ b }{ 3 }
C. h
D. -\frac{ h }{ 3 }
E. \frac{ h }{ 3 }

Answer 1

do u realize that means alpha and beta are roots of equation 3x^2-hx-b=0 ?

Answer 2

no

Answer 3

oh, if alpha and beta are roots of equation 3x^2-hx-b=0 ?
then u can write
\(3\alpha ^{2}-h \alpha-b=0 \\3\beta ^{2}-h \beta-b=0\)
here thats given, and since converse is also true....u can conclude that
alpha and beta are roots of equation 3x^2-hx-b=0
got that ?

Answer 4

I am especially weak in roots......

Answer 5

2. if α and β are the roots of the equation x^(2)-4x-3=0, then α^(2)+αβ+β^(2)=?
A. -13
B. 5
C. 13
D. 16
E. 19

3. Find the range of values of k such that the equation x^(2)+(k-2)x+1=0 has real roots.
A. k=4
B. 0<k<4
C. \[0 \le k \le4\]
D. k<0 or k>4
E. \[k \le 0 \ / k \ge 4\]

Answer 6

hey u got 1st answer ?

Answer 7

no.....I just type a few more questions

Answer 8

did u get how alpha and beta are roots of equation 3x^2-hx-b=0

Answer 9

h^(2)+12b ??

Answer 10

what is h^(2)+12b
i didn't get you.... i just asked you whether u got my explanation....

Answer 11

errrr...well, I can't get it...

Answer 12

\[3\alpha ^{2}-h \alpha-b=0---(1)\]
\[3\beta ^{2}-h \beta-b=0---(2)\]
(2)-(1),
\[3\alpha ^{2}-h \alpha+3\beta ^{2} -hb=0---(3)\]
may I do it like this?

Answer 13

i don't think that would help.....but let me see...

Answer 14

however, I cannot get the answer...

Answer 15

3a^2-3b^2 -h(a-b) = 0
(a-b) [3(a+b)-h]=0

Answer 16

yes! thats it

Answer 17

did u get what i've done ?

Answer 18

well....

3α^(2)−hα+3β^(2)−hb
=3[α^(2)+β^(2)]-h(α+β)

isn't it?

Answer 19

\(3\alpha ^{2}-h \alpha-3\beta ^{2} +hb=0---(3) \\ 3(\alpha^2-\beta)^2-h(\alpha-\beta)=0 \\ 3(\alpha-\beta)(\alpha+\beta)-h(\alpha-\beta)=0 \\ \text{factor out }\alpha -\beta\)

Answer 20

you subtracted incorrectly

Answer 21

do (2)-(1) again

Answer 22

what u get ?

Answer 23

3[α^(2)-β^(2)]-h(α-β)=0

Answer 24

yeah , now see my explanation above and factor out alpha - beta
what u get ?

Answer 25

3(α−β)(α+β)−h(α−β)=0
(α−β)(α+β)(3-h)=0

is it? I am weak in this part..

Answer 26

\(3\alpha ^{2}-h \alpha-3\beta ^{2} +hb=0---(3) \\ 3(\alpha^2-\beta)^2-h(\alpha-\beta)=0 \\ 3(\alpha-\beta)(\alpha+\beta)-h(\alpha-\beta)=0 \\ \text{factor out }\alpha -\beta \\ (\alpha -\beta)(3(\alpha+\beta)-h)=0\)

Answer 27

okay, got it

Answer 28

how come I got the answer like this: