Numbers can change the Perceptions. Part II

Algebra Level 5

x 2 + x 2 + 1729 + x 2 − x 2 + 1729 = 17.29 \sqrt{x^2+\sqrt{x^2+1729}}+\sqrt{x^2-\sqrt{x^2+1729}}=17.29

Find the number of real values of x x satisfying the above equation.

More than two Two Zero One

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Personal Data
Jun 6, 2015

The function on the L H S LHS is even so we'll only consider positive values of x x .

We can clearly see that it's increasing since both the first term and the second are increasing (the first is obvious, for the second we can notice that x 2 { x }^{ 2 } grows faster than x 2 + 1729 \sqrt { { x }^{ 2 }+1729 } ).

The function will be defined if x 2 − x 2 + 1729 ≥ 0 \sqrt { { x }^{ 2 }-\sqrt { { x }^{ 2 }+1729 } } \ge 0 or equivalently x 4 − x 2 − 1729 ≥ 0 { x }^{ 4 }-{ x }^{ 2 }-1729\ge 0 which has a solution x ≥ 1 + 6917 2 x\ge \sqrt { \frac { 1+\sqrt { 6917 } }{ 2 } } or x ≤ − 1 + 6917 2 x\le -\sqrt { \frac { 1+\sqrt { 6917 } }{ 2 } } .

We now that the function is increasing in its domain so min ⁡ { x 2 + x 2 + 1729 + x 2 − x 2 + 1729 } = 6917 + 1 < 17.29 \min { \left\{ \sqrt { { x }^{ 2 }+\sqrt { { x }^{ 2 }+1729 } } +\sqrt { { x }^{ 2 }-\sqrt { { x }^{ 2 }+1729 } } \right\} =\sqrt { \sqrt { 6917 } +1 } } <17.29 at x = 1 + 6917 2 x=\sqrt { \frac { 1+\sqrt { 6917 } }{ 2 } } .

Combining those facts we know that there are exactly two solutions to the equation.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

x 2 + x 2 + 1729 + x 2 − x 2 + 1729 = 17.29 ∴ x 2 + x 2 + 1729 + x 2 − x 2 + 1729 + 2 ∗ { x 2 + x 2 + 1729 ∗ x 2 − x 2 + 1729 } = 2 x 2 + 2 ∗ { x 4 − x 2 − 1729 } = 17.2 9 2 ⟹ x 4 − x 2 − 1729 = ( 17.29 ) 4 16 + x 4 − 2 ∗ x 2 ∗ ( 17.29 ) 2 4 ∴ x 2 ( 1 + 2 ∗ ( 17.29 ) 2 4 ) = ( 17.29 ) 2 4 + 1729 x = 11.987. ∴ x 2 − x 2 + 1729 = 100.417 > 0 So x is real and has two values, +x and -x \sqrt{x^2+\sqrt{x^2+1729}}+\sqrt{x^2-\sqrt{x^2+1729}}=17.29\\ \therefore~ x^2+\sqrt{x^2+1729}+x^2-\sqrt{x^2+1729} \\+2*\left\{\sqrt{x^2+\sqrt{x^2+1729}}*\sqrt{x^2-\sqrt{x^2+1729}} \right \} \\=2x^2+2*\left\{\sqrt{x^4- x^2-1729} \right \}=17.29^2 \\ \implies~ x^4- x^2-1729=\dfrac{(17.29)^4} {16}+x^4-2*x^2*\dfrac{(17.29)^2} 4 \\\therefore~x^2(1+2*\dfrac{(17.29)^2} 4 )=\dfrac{(17.29)^2} 4 +1729 \\ x=11.987. ~~\therefore~\sqrt{x^2-\sqrt{x^2+1729}}=100.417>0\\\text{So x is real and has two values, +x and -x }

Niranjan Khanderia - 5 years, 11 months ago
Chew-Seong Cheong
Jul 10, 2015

x 2 − x 2 + 1729 \sqrt{x^2-\sqrt{x^2+1729}} is always not real. The equation has no solution. Therefore the answer is Zero \boxed{\text{Zero}} .

2 Helpful 1 Interesting 1 Brilliant 1 Confused

Moderator note:

Thanks for pointing that out!

why ??? is this always non real

Kaustubh Miglani - 5 years, 11 months ago

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Sorry, I have got it wrong. I have informed Calvin. I have a chemo brain.

Chew-Seong Cheong - 5 years, 11 months ago

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no u are a great person.please be my tutor

Kaustubh Miglani - 5 years, 11 months ago

I too got it wrong. Please do not thing it is because of chemo. You have always solved many problem I have not solved. I had no major illness except asthma a few years back. Still you do better. At time it is natural to miss something. You are exceptional.

Niranjan Khanderia - 5 years, 11 months ago

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@Niranjan Khanderia – Thanks Niranjan and Asdfghjk for your remarks. But the twice nine-week chemotherapy I had really affect the brain. I can't do integration by part like when I was in high school.

Chew-Seong Cheong - 5 years, 11 months ago

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