Algebra Of Square Roots

Algebra Level 3

Let x 1 x_1 , x 2 x_2 , x 3 x_3 , x 4 x_4 , and x 5 x_5 be real numbers satisfying

K = x 1 1 + 2 x 2 4 + 3 x 3 9 + 4 x 4 16 + 5 x 5 25 = x 1 + x 2 + x 3 + x 4 + x 5 2 K = \sqrt{x_1 - 1} + 2\sqrt{x_2 - 4} + 3\sqrt{x_3 - 9} + 4 \sqrt{x_4 - 16} + 5\sqrt{x_5 - 25} = \dfrac{x_1 + x_2 + x_3 + x_4 + x_5}2

Find the value of K 2 . K^2 .

1225 2025 3025 5625

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1 solution

From the given equation, ( x 1 2 x 1 1 x_1-2\sqrt{x_1-1} ) + ( x 2 4 x 2 4 x_2-4\sqrt{x_2-4} ) + ( x 3 6 x 3 9 x_3-6\sqrt{x_3-9} ) + ( x 4 8 x 4 16 x_4-8\sqrt{x_4-16} ) + ( x 5 10 x 5 25 x_5-10\sqrt{x_5-25} ) =0

( x 1 1 1 ) 2 (\sqrt{x_1-1}-1)^{2} + ( x 2 4 2 ) 2 (\sqrt{x_2-4}-2)^{2} + ( x 3 9 3 ) 2 (\sqrt{x_3-9}-3)^{2} + ( x 4 16 4 ) 2 (\sqrt{x_4-16}-4)^{2} = ( x 5 25 5 ) 2 (\sqrt{x_5-25}-5)^{2} =0

So, x 1 x_1 =2, x 2 x_2 =8, x 3 x_3 =18, x 4 x_4 =32, x 5 x_5 =50.

x 1 + x 2 + x 3 + x 4 + x 5 2 \frac{x_1+x_2+x_3+x_4+x_5}{2} =55, 5 5 2 55^{2} =3025 .

Hope this helps. :) Special Thanks to Chris Lewis for correcting my mistake.

Hi - nice problem! I think you have a typo in your solution, though; on the second line, your equation says ( x 5 25 1 ) 2 = 0 \left(\sqrt{x_5-25}-1\right)^2=0 from which you infer x 5 = 50 x_5=50 . Could you explain how you get from your first step to your second?

Chris Lewis - 2 months, 4 weeks ago

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The real equation is actually

( x 1 1 1 ) 2 + ( x 2 4 2 ) 2 + ( x 3 9 3 ) 2 + ( x 4 16 4 ) 2 + ( x 5 25 5 ) 2 = 0 (\sqrt{x_1-1}-1)^2+ (\sqrt{x_2-4}-2)^2+ (\sqrt{x_3-9}-3)^2+ (\sqrt{x_4-16}-4)^2+ (\sqrt{x_5-25}-5)^2=0

Since squares can’t be negative, all of them have to be equal zero

Jason Gomez - 2 months, 4 weeks ago

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Yup, I agree that's what it should say. The solution needs a few edits.

Chris Lewis - 2 months, 4 weeks ago

Oh extremely sorry, those are gonna be 1, 2,3,4,5....

Baibhab Chakraborty - 2 months, 3 weeks ago

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