Menu Close

solve-x-2-x-2-x-2-x-1-x-

Tinku Tara 0 Comments
Pin




Question Number 171402 by mnjuly1970 last updated on 14/Jun/22
solve... ⌊x^( 2) ⌋+⌊x⌋^( 2) = x^( 2) +x +1 x=?
solve…⌊x2⌋+⌊x⌋2=x2+x+1x=?
Answered by floor(10²Eta[1]) last updated on 14/Jun/22
⌊x^2 ⌋=a∈Z⇒x^2 =a+α, 0≤α<1 ⌊x⌋=b∈Z⇒x=b+β, 0≤β<1 a+b^2 =a+α+b+β+1 ⇒b^2 −b=α+β+1⇒1≤b^2 −b<3 ⇒b^2 −b∈{1,2}⇒b^2 −b=2⇒b=2 or b=−1 1 case: b=2 ⌊x⌋=2, x=2+α⇒x^2 =4+4α+α^2 ⇒4≤x^2 <9 x^2 ∈[4,5)∴⌊x^2 ⌋=4 4+2^2 =x^2 +x+1⇒x^2 +x−7=0 ⇒x=((−1+(√(29)))/2) x^2 ∈[5,6)∴⌊x^2 ⌋=5 5+2^2 =x^2 +x+1⇒x^2 +x−8=0 ⇒x=((−1+(√(33)))/2) x^2 ∈[6,7)∴⌊x^2 ⌋=6 6+2^2 =x^2 +x+1⇒x^2 +x−9=0 x=((−1+(√(37)))/2) x^2 ∈[7,8)∴⌊x^2 ⌋=7 7+2^2 =x^2 +x+1⇒x^2 +x−10=0 ⇒x=((−1+(√(41)))/2) x^2 ∈[7,8)∴⌊x^2 ⌋=8 ⇒x=((−1+(√(45)))/2) 2 case: b=−1 ⌊x⌋=−1, x=−1+α⇒x^2 =1−2α+α^2 ⇒0<x^2 ≤1 x^2 ∈(0,1)∴⌊x^2 ⌋=0 0+(−1)^2 =x^2 +x+1⇒x^2 +x=0 ⇒x=−1⇒x^2 ∉(0,1) (dont work) x^2 =1∴⌊x^2 ⌋=1 1+(−1)^2 =x^2 +x+1⇒x^2 +x−1=0 x=((−1−(√5))/2)≤−1(dont work) solutions: x=((−1+(√(25+4n)))/2), 1≤n≤5, n∈N
⌊x2⌋=a∈Z⇒x2=a+α,0⩽α<1⌊x⌋=b∈Z⇒x=b+β,0⩽β<1a+b2=a+α+b+β+1⇒b2−b=α+β+1⇒1⩽b2−b<3⇒b2−b∈{1,2}⇒b2−b=2⇒b=2orb=−11case:b=2⌊x⌋=2,x=2+α⇒x2=4+4α+α2⇒4⩽x2<9x2∈[4,5)∴⌊x2⌋=44+22=x2+x+1⇒x2+x−7=0⇒x=−1+292x2∈[5,6)∴⌊x2⌋=55+22=x2+x+1⇒x2+x−8=0⇒x=−1+332x2∈[6,7)∴⌊x2⌋=66+22=x2+x+1⇒x2+x−9=0x=−1+372x2∈[7,8)∴⌊x2⌋=77+22=x2+x+1⇒x2+x−10=0⇒x=−1+412x2∈[7,8)∴⌊x2⌋=8⇒x=−1+4522case:b=−1⌊x⌋=−1,x=−1+α⇒x2=1−2α+α2⇒0<x2⩽1x2∈(0,1)∴⌊x2⌋=00+(−1)2=x2+x+1⇒x2+x=0⇒x=−1⇒x2∉(0,1)(dontwork)x2=1∴⌊x2⌋=11+(−1)2=x2+x+1⇒x2+x−1=0x=−1−52⩽−1(dontwork)solutions:x=−1+25+4n2,1⩽n⩽5,n∈N
Answered by mr W last updated on 14/Jun/22
x=n+f ⌊x⌋^2 =n^2 ⌊x^2 ⌋=n^2 +⌊(2n+1)f⌋ x^2 +x+1=n^2 +n+1+(2n+1)f+f^2 2n^2 +⌊(2n+1)f⌋=n^2 +n+1+(2n+1)f+f^2 n^2 −n−1=(2n+1)f+f^2 −⌊(2n+1)f⌋ n^2 −n−1={(2n+1)f}+f^2 ≥0 ⇒n≤−1 or n≥2 n^2 −n−1={(2n+1)f}+f^2 <2 ⇒−1≤n≤2 ⇒n=−1 or n=2 with n=−1, i.e. −1≤x<0: 1+1=x^2 +x+1 x^2 +x−1=0 ⇒x=((−1−(√5))/2) <−1 rejected. with n=2, i.e. 2≤x<3: ⌊x^2 ⌋=k with 4≤k≤8 k+4=x^2 +x+1 x^2 +x−(k+3)=0 ⇒x=((−1+(√(13+4k)))/2) ✓ i.e. x=((−1+(√(29)))/2), ((−1+(√(33)))/2), ((−1+(√(37)))/2), ((−1+(√(41)))/2), ((−1+(√(45)))/2)
x=n+f⌊x⌋2=n2⌊x2⌋=n2+⌊(2n+1)f⌋x2+x+1=n2+n+1+(2n+1)f+f22n2+⌊(2n+1)f⌋=n2+n+1+(2n+1)f+f2n2−n−1=(2n+1)f+f2−⌊(2n+1)f⌋n2−n−1={(2n+1)f}+f2⩾0⇒n⩽−1orn⩾2n2−n−1={(2n+1)f}+f2<2⇒−1⩽n⩽2⇒n=−1orn=2withn=−1,i.e.−1⩽x<0:1+1=x2+x+1x2+x−1=0⇒x=−1−52<−1rejected.withn=2,i.e.2⩽x<3:⌊x2⌋=kwith4⩽k⩽8k+4=x2+x+1x2+x−(k+3)=0⇒x=−1+13+4k2✓i.e.x=−1+292,−1+332,−1+372,−1+412,−1+452
Commented by floor(10²Eta[1]) last updated on 14/Jun/22
if x=−1⇒x^2 =1 ⌊x^2 ⌋+⌊x⌋^2 =2≠x^2 +x+1=1
ifx=−1⇒x2=1⌊x2⌋+⌊x⌋2=2≠x2+x+1=1
Commented by mr W last updated on 14/Jun/22
i have rejected x=−1.
ihaverejectedx=−1.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *