Q. 145.0( 1 Vote )

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Answer :


(i) Direct method: x3 + 4x = 0 or x(x2 + 4) = 0
Now x2 + 4 ≠ 0 and x ∈ R and hence x = 0.

(ii) Method of Contradiction: Let x ≠ 0 and let x = p, p ∈ R is a root of x3 + 4x = 0. Therefore p3 + 4p = 0
(or) p(p2 + 4) = 0 as p = 0 Thus p2 + 4 = 0 which is not possible. Therefore, our supposition is wrong. Hence p = 0 or x = 0
 
(iii) Contra positive method: p is not true

Let x = 0 is not true Let x = p ≠ 0

Therefore p3 + 4p = 0

P being the root of x2 + 4 = 0

(or) p(p2 + 4) = 0

Now p = 0, also (p2 + 4) = 0

Which implies p(p2 + 4) = 0

Now p = 0, also p2 + 4 = 0

p(p2 + 4) ≠ 0 if p is not true.

Hence x = 0 is the root of x3 + 4x = 0

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