Maths problem, how to do,thanks?

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Method 1 (More standard and students should learn this)
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Given that 3x² - 17x - 28 = 0 has roots α and β,
we know α + β = 17/3 and αβ = -28/3.

For the quadratic equation to be found,
sum of roots is 3α + 3β = 3(α + β) = 3 × 17/3 = 17, and
product of roots is (3α)(3β) = 9αβ = 9 × (-28/3) = -84.

Therefore, the required equation can be
x² - 17x - 84 = 0
That is, (A).


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Method 2
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It is given that the equation 3x² - 17x - 28 = 0 ...(*) has roots α and β in x.

An equation with roots 3α and 3β in x is the same as an equation with roots α and β in x/3.

Replacing x by x/3 in (*) gives
3(x/3)² - 17(x/3) - 28 = 0
3(x²/9) - 17x/3 - 28 = 0
x² - 17x - 84 = 0 is the required equation.


Remark:
To elaborate more on Method 2 (can be applied to general polynomial), suppose a₁, a₂, ..., aₙ are the roots of p(x) = 0 where p(x) is a polynomial.

Then a polynomial equation having roots g(a₁), g(a₂), ..., g(aₙ) where g is invertible can be p( g⁻¹(x) ) = 0.