Question

if a(alpha) and b(beta) are roots of equation xsquare-7x+7=0 then find(1/a + 1/b)ab

Answer 1

Note that:\[\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\cdot \alpha\beta=\alpha + \beta\]So what you are looking for is the sum of the roots.

Also note that if you have a quadratic equation: \[ax^2+bx+c=0\]then the sum of the roots is:\[-\frac{b}{a}\]In your case, a = 1 and b = -7, so the sum of the roots is 7.

Answer 2

I posted your steps

Answer 3

using completing the square, since the given trinomial is not a perfect trinomial, let's make it a perfect one. .

equation: \[x ^{2}-7x+7=0\]
find \[(1/\alpha + 1/\beta)\alpha \beta\]
make the given equation a perfect trinomial:

\[x ^{2}-7x+7=0\]\[x ^{2}-7x+7+5=5\]\[x ^{2}-7x+12=5\]\[(x-3)(x-4)=5\]
x-3=5
x=8

x-4=5
x=9

therefore roots are (8,9)(a,b)

substitute:

\[(1/a+1/b)ab\](1/8+1/9)(8)(9)
9+8


17 ; final answer

Answer 4

I like how he just posts and leave.

Answer 5

That is incorrect =/ when you got to:\[(x-3)(x-4)=5\] It is incorrect to say x = 8 or x = 9. You can test this by plugging in 8 or 9:\[(8-3)(8-4)=5\cdot 4=20\]

Answer 6

meh, its whatever.
what did you mean by "I posted your steps"?

Answer 7

i don't think there is flaw on my solution???

Answer 8

When you have:\[(x-3)\cdot (x-4)=0\]it is true that either x-3 = 0, or x-4 = 0. If you have:\[(x-3)\cdot (x-4) = 5\] It is not true that x-3 = 5 or x - 4 = 5.

Answer 9

When 2 numbers multiply to make 0, it is true that one of them must be 0.

If 2 numbers multiply to make 5, they could be anything, like 1 and 5, or 2.5 and 2, etc.

Answer 10

you might question why i got 5, well, that's part of the method of completing the square..

Answer 11

I see where you got the 5 from. What im saying is wrong is this step:
\[(x-3)(x-4)=5\]\[x-3=5\Longrightarrow x=8\]\[x-4=5\Longrightarrow x=9\] You are claiming that the roots of the polynomial \[x^2-7x+7=0\]are x = 8 and x = 9, but that is false. We can tell by plugging in the values:
\[(8)^2-7(8)+7=15\neq 0 \]\[(9)^2-7(9)+7=25\neq 0\]You have a clever idea, but the step i posted above is incorrect. So your answer is also incorrect.

Answer 12

yeah.. i got your point!!

Answer 13

hey guys i m confused plz tell me which is correct answer finally?