JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 7)
If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is
$$-$$ 81
$$-$$ 300
100
144
Explanation
81x2 + kx + 256 = 0 ; x = $$\alpha $$, $$\alpha $$3
$$ \Rightarrow $$ $$\alpha $$4 = $${{256} \over {81}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$ \pm $$ $${{4} \over {3}}$$
Now $$-$$ $${k \over {81}}$$ = $$\alpha $$ + $$\alpha $$3 = $$ \pm $$ $${{100} \over {27}}$$
$$ \Rightarrow $$ k = $$ \pm $$300
$$ \Rightarrow $$ $$\alpha $$4 = $${{256} \over {81}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$ \pm $$ $${{4} \over {3}}$$
Now $$-$$ $${k \over {81}}$$ = $$\alpha $$ + $$\alpha $$3 = $$ \pm $$ $${{100} \over {27}}$$
$$ \Rightarrow $$ k = $$ \pm $$300
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