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If a/k,a,ak are the roots of x³-px²+qx-r=0 then k= |
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Answer» Answer: => a = ∛rStep-by-step explanation: 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 = ∛rHope this is HELPFUL to you Pls MARK as brainliest |
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