This question looks harder than it appears (quadratic)

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You don't need to solve A and B.

You should know facts about the roots of polynomials.

$A+B=-\frac{b}{a}$

$AB=\frac{c}{a}$

So write down A+B and AB.

Now consider $\frac{2}{A}+\frac{2}{B}$

Can you rewrite this using A+B and AB?

Finally consider $\frac{2}{A} \times \frac{2}{B}$

Rewrite this using AB.

Now for your final answer. The quadratic is:

$x^2 -(\frac{2}{A}+\frac{2}{B})x+ (\frac{2}{A} \times \frac{2}{B})$

You will need to multiply the whole equation by 3 to obtain integer coefficients.

ok, this is what I got:

$A+B= -3,AB=-6$

$\frac{2}{A}+\frac{2}{B}=\frac{2B+2A}{AB}=\frac{2(A+B)}{AB}=\frac{2(-3)}{(-6)}=1$

$\frac{2}{A} \times \frac{2}{B}=\frac{4}{AB}=\frac{4}{(-6)}=\frac{2}{(-3)}$

$x^2+x-\frac{2}{3}=0$

$3x^2+3x-2=0$

correct?

Look at the coefficient of x.

Remember the sum of the roots is $=- \frac{b}{a}$

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Look at the coefficient of x.

Remember the sum of the roots is $=-\frac{b}{a}$

what do you mean? b = 3 a = 1 so -b/a = -3?

Do you remember doing this in A Level Further Maths?

I'm talking about the second equation (the one you are forming). Your sign is wrong for the coefficient of x.