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Assertion :If $$a, b,c \in Q$$ and $$2^{\dfrac{1}{3}}$$ is a root of $$ax^2+bx +c=0, then \ a=b=c=0$$. Reason: A polynomial equation with rational coefficients cannot have irrational roots


A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Assertion is incorrect but Reason is correct
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Solution

The correct option is C Assertion is correct but Reason is incorrect
$$a,b,c\epsilon Q$$
$${ 2 }^{ 1/3 }$$ is root of $$a{ x }^{ 2 }+bx+c=0$$
$$\therefore a\left( { 2 }^{ 2/3 } \right) +b\left( { 2 }^{ 1/3 } \right) +C=0\rightarrow 1$$
a,b,c are rational, $${ 2 }^{ 2/3 },{ 2 }^{ 1/3 }$$ are irrational
$$1$$ is true only if $$a=b=c=0$$
A polynomial of degree 'n' can have irrational roots with power $$\ge \cfrac { 1 }{ n } $$ (i.e., $$\cfrac { 1 }{ { n }^{ th } } $$ irrational power numbers)
$$\therefore $$ For a quadratic equation $$\left( { a }_{ 1 }^{ 1/2 }+{ b }_{ 1 }^{ 1/2 }+{ c }_{ 1 }^{ 1/2 }....+{ z }_{ 1 }^{ 1/2 } \right) $$
is possible as a root but $$\left( { a }_{ 2 }^{ 1/3 }+{ b }_{ 2 }^{ 1/3 }+....+{ z }_{ 2 }^{ 1/3 } \right) $$ is not possible as a root
$$\left( \because \cfrac { 1 }{ 3 } <\cfrac { 1 }{ 2 }  \right) $$

Mathematics

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