Amazing Roots

Algebra Level 4

x + 1 x 1 = 4 x 1 \large \sqrt {x+1} - \sqrt {x-1} = \sqrt {4x-1}

Find the number of real roots that satisfy the equation above.

0 2 1 None of these choices

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2 solutions

Otto Bretscher
Sep 28, 2015

In the domain, x 1 x\geq{1} , we have x + 1 < 4 x 1 \sqrt{x+1}<\sqrt{4x-1} , so L H S < R H S LHS < RHS

Moderator note:

Nice inequality comparison that gives the result immediately without having to square terms.

5/4 is satisfying

Saksham Rastogi - 5 years, 8 months ago

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No. Its not. As Mr. Otto Bretscher says L H S = 1 , R H S = 2 LHS = 1, RHS = 2

Vishal Yadav - 5 years, 8 months ago

see the MINUS sign, not plus sign, between the term x + 1 \sqrt{x+1} and x 1 \sqrt{x-1} .

Gian Sanjaya - 5 years, 8 months ago

Sir, can you explain me how did you come to the conclusion that x + 1 \sqrt{x+1} < < 4 x 1 \sqrt {4x-1} . And if it is so then how L H S < R H S LHS < RHS signify that the above equation has no real roots. Thanks.

Vishal Yadav - 5 years, 8 months ago

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We have x 1 x\geq{1} so 3 x > 2 3x>2 so 4 x 1 > x + 1 4x-1>x+1 so 4 x 1 > x + 1 \sqrt{4x-1}>\sqrt{x+1} .

If L H S < R H S LHS<RHS , then they aren't equal. ;)

Otto Bretscher - 5 years, 8 months ago

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Sir, you took x 1 x \geq 1 . But what about the domain x < 1 x < 1 . The solutions will still remain real.(correct me if I am wrong)

Vishal Yadav - 5 years, 8 months ago

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@Vishal Yadav The problem asks for real roots, so, the domain is x 1 x\geq 1 because of the term x 1 \sqrt{x-1}

Otto Bretscher - 5 years, 8 months ago

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@Otto Bretscher Sir, I think that real roots mean that x R x\in R and does not include i i as a + b i a + bi . This definition holds true for x < 0 x < 0

Vishal Yadav - 5 years, 8 months ago

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@Vishal Yadav There clearly are no solutions x < 1 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 x\geq 1

Otto Bretscher - 5 years, 8 months ago

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@Otto Bretscher Thanks Sir. You gave me a time saving solution.

Vishal Yadav - 5 years, 8 months ago

draw diff. graphs ! simply the best method which comes only by practice!

A Former Brilliant Member - 4 years, 9 months ago
Shaheryar Waris
Oct 7, 2015

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??

Sparsh Setia - 4 years, 6 months ago

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