Kwadrat sumy trzech kolejnych liczb naturalnych jest o 1534 większy od sumy ich kwadratów. Wyznacz te liczby.
(n+n+1+n+2)²=n²+(n+1)²+(n+2)²+1534
(3n+3)²=n²+n²+2n+1+n²+4n+4+1534
9n²+18n+9=3n²+6n+1539
6n²+12n-1530=0
n²+2n-255=0
Δ=2²-4*1*(-255)
Δ=4+1020
Δ=1024
√Δ=32
n₁=(-2-32)/(2*1)
n₁=-34/2
n₁=-17 <-- odpada bo n∉N
n₂=(-2+32)/(2*1)
n₂=30/2
n₂=15
n=15
n+1=16
n+2=17
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(n+n+1+n+2)²=n²+(n+1)²+(n+2)²+1534
(3n+3)²=n²+n²+2n+1+n²+4n+4+1534
9n²+18n+9=3n²+6n+1539
6n²+12n-1530=0
n²+2n-255=0
Δ=2²-4*1*(-255)
Δ=4+1020
Δ=1024
√Δ=32
n₁=(-2-32)/(2*1)
n₁=-34/2
n₁=-17 <-- odpada bo n∉N
n₂=(-2+32)/(2*1)
n₂=30/2
n₂=15
n=15
n+1=16
n+2=17