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I have to find the roots to this equation

$$|x^2-[x]| =1$$
Where [x] is the greatest integer function and |a| is the modulus of a.

I don’t know how to go about it, hit and trial gave me nothing, help me out.

$|x^2-[x]|=x^2-[x]=1because x^2ge[x]forall xinBbb R$

$[x]=x-{x}$, where ${cdot}$ denotes the fractional part of $x$

$implies x^2-x+{x}-1=0$

$impliesdisplaystyle x=frac{1pmsqrt{5-4{x}}}{2}$

Now, $0le{x}<1implies1<5-4{x}le5implies xinbig(1,frac{1+sqrt5}2big]cupbig[frac{1-sqrt5}2,0big)$

In $big[frac{1-sqrt5}2,0big)$, $[x]=-1implies x^2=[x]+1=0implies x=0notinbig[frac{1-sqrt5}2,0big)$

In $big(1,frac{1+sqrt5}2big],[x]=1implies x^2=2implies x=sqrt2$

The final answer is $x=sqrt2$.

Here is how I solved it:

Consider x in (k, k+1]
x = k+r
|x²-ceil(x)| = |(k+r)²-(k+1)| = 1

Case1: (k+r)²-(k+1) >= 0
|(k+r)²-(k+1)| = (k+r)²-k-1 = 1
(k²+2kr+r²)-k = 2
1r² +2kr +k²-k-2 = 0
r in (-2k +/- sqrt(4k²-4(k²-k-2))) / 2
= -k +/- sqrt(k²-(k²-k-2))
= -k +/- sqrt(k²-k²+k+2)
= -k +/- sqrt(k+2)
= {-k + sqrt(k+2), -k - sqrt(k+2)} = {r1, r2}
For real roots, k must be in {-2, -1, ...}
r must be within (0, 1]
0 < r1:
-k + sqrt(k+2) > 0
sqrt(k+2) > k
true for k <= 0, consider k > 0
k+2 > k²
1k² - 1k - 2 < 0
(1 +/- sqrt(1 - 4*(-2))) / 2 = (1 +/- 3) / 2 = {-1, 2}
k > 0 and 1k² - 1k - 2 < 0 is false for k → inf
0 < k < 2
k = 1
k in {-2, -1, 0, 1}
r1 <= 1:
-k + sqrt(k+2) <= 1
sqrt(k+2) <= 1+k
false for k < -1
k+2 <= (1+k)² = k²+2k+1
0 <= k² + k - 1
0 <= (-1)² + (-1) - 1 = -1, false
0 <= (0)² + (0) - 1 = -1, false
0 <= (1)² + (1) - 1 = 1, true
k = 1
r1 = -k + sqrt(k+2) = -1 + sqrt(1+2) = sqrt(3) - 1
root candidate: 1 + r1 = 1 + sqrt(3) - 1 = sqrt(3)
0 < r2:
-k - sqrt(k+2) > 0
k < -sqrt(k+2), false for k >= 0
-1 < -sqrt(-1+2), 1 > sqrt(1), false
-2 < -sqrt(-2+2), 2 > sqrt(0), true
k = -2
r2 <= 1:
r2 = -k - sqrt(k+2) = 2 - sqrt(-2+2) = 2, false

Case2: (k+r)²-(k+1) < 0
|(k+r)²-(k+1)| = -(k+r)²+k+1 = 1
-(k²+2kr+r²)+k = 0
k²+2kr+r²-k = 0
1r² + 2kr + k²-k = 0
r in (-2k +/- sqrt(4k² - 4(k²-k))) / 2
= -k +/- sqrt(k² - (k²-k))
= -k +/- sqrt(k)
= {-k - sqrt(k), -k + sqrt(k)}
For real roots, k must be in {0, 1, ...}
r must be within (0, 1]
0 < r1 = -k - sqrt(k) is false for those k
0 < r2:
0 < -k + sqrt(k), k < sqrt(k), k² < k, k < 0, false
No root candidates for Case2

The only root found is sqrt(3) when Case1 pertains, a simple test shows that
this is indeed the root:
|sqrt(3)²-ceil(sqrt(3))| = 1
|3-2| = 1
1 = 1