Real roots !!

Algebra Level 2

has which real roots ???

2,17 2,5 10,17 5,10

Shivram Badhe by Shivram Badhe , 21, India

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1 solution

This probably isn't the best solution but I solved it this way.First put 2 in the equation and evaluate to see if the result is 1.It isn't.As both roots are supposed to satisfy the equation,so 1st and 4th option are incorrect(because 2 is not a solution).Try the next option which is 5,10.Evaluate the expression by substituting x x with both of these values and you will get that the R.H.S is equal to 1 so answer is 5 , 10 5,10 .I think verifying the options is easier than solving in this case

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Ultimately even I came down to trial and error, but a little simplification before that came handy.

x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1

( x − 1 ) − 2 ( x − 1 ) ( 2 ) + 4 + ( x − 1 ) − 2 ( x − 1 ) ( 3 ) + 9 = 1 \sqrt{(x-1)-2\left(\sqrt{x-1}\right)(2)+4}+\sqrt{(x-1)-2\left(\sqrt{x-1}\right)(3)+9}=1

( x − 1 ) 2 − 2 ( x − 1 ) ( 2 ) + ( 2 ) 2 + ( x − 1 ) 2 − 2 ( x − 1 ) ( 3 ) + ( 3 ) 2 = 1 \sqrt{\left(\sqrt{x-1}\right)^{2}-2\left(\sqrt{x-1}\right)(2)+(2)^{2}}+\sqrt{\left(\sqrt{x-1}\right)^{2}-2\left(\sqrt{x-1}\right)(3)+(3)^{2}}=1

( x − 1 − 2 ) 2 + ( x − 1 − 3 ) 2 = 1 \sqrt{\left(\sqrt{x-1}-2\right)^{2}}+\sqrt{\left(\sqrt{x-1}-3\right)^{2}}=1

∣ x − 1 − 2 ∣ + ∣ x − 1 − 3 ∣ = 1 \lvert\sqrt{x-1}-2\rvert+\lvert\sqrt{x-1}-3\rvert=1

This makes trial and error easier.

Omkar Kulkarni - 6 years, 2 months ago

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