x + 1 − x − 1 = 4 x − 1
Find the number of real roots that satisfy the equation above.
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Nice inequality comparison that gives the result immediately without having to square terms.
5/4 is satisfying
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No. Its not. As Mr. Otto Bretscher says L H S = 1 , R H S = 2
see the MINUS sign, not plus sign, between the term x + 1 and x − 1 .
Sir, can you explain me how did you come to the conclusion that x + 1 < 4 x − 1 . And if it is so then how L H S < R H S signify that the above equation has no real roots. Thanks.
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We have x ≥ 1 so 3 x > 2 so 4 x − 1 > x + 1 so 4 x − 1 > x + 1 .
If L H S < R H S , then they aren't equal. ;)
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Sir, you took x ≥ 1 . But what about the domain x < 1 . The solutions will still remain real.(correct me if I am wrong)
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@Vishal Yadav – The problem asks for real roots, so, the domain is x ≥ 1 because of the term x − 1
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@Otto Bretscher – Sir, I think that real roots mean that x ∈ R and does not include i as a + b i . This definition holds true for x < 0
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@Vishal Yadav – There clearly are no solutions x < 1 since the imaginary part of the LHS is negative while the imaginary part of the RHS is nonnegative. Thus we can focus on the domain x ≥ 1
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@Otto Bretscher – Thanks Sir. You gave me a time saving solution.
draw diff. graphs ! simply the best method which comes only by practice!
put 2 or three real numbers they will not satisfy the equation so there are 0 real roots of this equation as roots always satisfy the equation...
you dumb. Is this the way you solve problems??
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In the domain, x ≥ 1 , we have x + 1 < 4 x − 1 , so L H S < R H S