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Algebra Difficulty 7.4 National Olympiad, round 2 Find the answer

Determine each real root of
x4−(2⋅1010+1)x2−x+1020+1010−1=0x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0
correct to four decimal places.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The equation can be re-written as (x+10 5) 2(x-10 5) 2 -(x+10 5)(x-10 5) -x-1=0.\text{(x+10 5) 2(x-10 5) 2 -(x+10 5)(x-10 5) -x-1=0.}
We first prove that the equation has no negative roots.
Let x≤0.x\le 0. The equation above can be further re-arranged as [(x+105)(x−105)+1][(x+105)(x−105)−2]=x−1.\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*} The right hand side of the equation is negative. Therefore [(x+105)(x−105)+1][(x+105)(x−105)−2)]<0,[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]<0, and we have −1<(x+105)(x−105)<2.-1<(x+10^5)(x-10^5) <2. Then the left hand side of the equation is bounded by ∣[(x+105)(x−105)+1][(x+105)(x−105)−2]∣≤3×3.|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3. However, since ∣(x+105)(x−105)∣≤2|(x+10^5)(x-10^5)|\le 2 and x<0,x<0, it follows that ∣x+105∣<2∣x−105∣<2×10−5|x+10^5| <\frac{2}{|x-10^5|}<2\times 10^{-5} for negative x.x. Then x<2×10−5−105.x<2\times 10^{-5}-10^5. The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.
Now let x>0.x>0. When x=105,x=10^5, the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of 10510^5 , as its leading coefficient is positive. We will prove that x=105x=10^5 is a good approximation of the roots (within 10−210^{-2} ). In fact, we can solve the "quadratic" equation (1) for (x+105)(x−105)(x+10^5)(x-10^5) : (x+105)(x−105)=1±1+4(x+1)2.(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}. Then x−105=1±1+4(x+1)2(x+105).x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}. Easy to see that ∣x−105∣<1|x-10^5| <1 for positve x.x. Therefore, 105−1<x<105+1.10^5-1<x<10^5+1. Then ∣x−105∣=∣1±1+4(x+1)2(x+105)∣≤∣12(x+105)∣+∣1+4(x+1)2(x+105)∣≤12(105−1+105)+1+4(105+1+1)2(105−1+105)<10−2.\begin{align*} |x-10^5|&=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\ &<10^{-2}. \end{align*}
Let x1x_1 be a root of the equation with x1<105.x_1<10^5. Then 0<105−x1<10−20<10^5-x_1<10^{-2} and x1−105=1−1+4(x1+1)2(x1+105).x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}. An aproximation of x1x_1 is defined as follows: x~1=105+1−1+4(105+1)2(105+105).\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}. We check the error of the estimate: ∣x~1−x1∣=∣1−1+4(105+1)2(105+105)−1−1+4(x1+1)2(x1+105)∣≤∣12(105+105)−12(x1+105)∣+∣1+4(105+1)2(105+105)−1+4(x1+1)2(x1+105)∣.\begin{align*} |\tilde{x}_1-x_1|&=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\ &\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |. \end{align*}
The first absolute value ∣12(105+105)−12(x1+105)∣=∣x1−105∣2(105+105)(x1+105)<10−12.\left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}<10^{-12}.
The second absolute value ∣1+4(105+1)2(105+105)−1+4(x1+1)2(x1+105)∣≤∣1+4(105+1)2(105+105)−1+4(x1+1)2(105+105)∣+∣1+4(x1+1)2(105+105)−1+4(x1+1)2(x1+105)∣≤10−7+10−9,\begin{align*} &\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right |\\ &\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\ &\le 10^{-7}+10^{-9}, \end{align*} through a rationalized numerator.Therefore ∣x~1−x1∣≤10−6.|\tilde{x}_1-x_1|\le 10^{-6}.
For a real root x2x_2 with x2>105,x_2>10^5, we choose x~2=105+1+1+4(105+1)2(105+105).\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}. We can similarly prove it has the desired approximation.

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