Question

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if a/k, a, ak are the roots of the equation x^2-px^2+qx-r=0 then a =​

Answer 1

When a, b, c are the roots of a standard cubic equation ax powerof3 + bx powerof3 + cx + d, then the multiplication of the roots = -d/a

Sum of roots = -b/a

So, multiplication of Roots = a powerof3

a powerof3 = r

So, a = r1/3

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Answer 2

p(x) = x³-px²+qx-r = 0

Since  a/k, a, ak are the roots of the equation,

(x-a/k) , (x-a) and (x-ak) are the factors of the equation

Hence

p(x) = (x-a/k)(x-a)(x-ak) = x³-px²+qx-r

(x-a/k)(x²-x(a+ak)+a²k) = x³-px²+qx-r

x³ - x²(a+ak) + xa²k - x²a/k + xa/k(a+ak) -a³ = x³-px²+qx-r

x³ + x²(-a-ak-a/k) + x(a²/k + a²) - a³ = x³-px²+qx-r

x²(-a-ak-a/k) + x(a²/k + a²) - a³ = -px²+qx-r

Hence on comparing both the equations

-a-ak-a/k = -p     => a(1+k+1/k) = p     => a = p/(1+k+1/k)

a²/k + a² = q       => a²(1/k+1) = q       => a = √[ q / (1/k+1) ]

-a³ = -r                                                  => a = ∛r

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