Question

Suma kwadratów pięciu kolejnych liczb całkowitych jest równa 135.Wyznacz te liczby.

Answer 1

N²+(n+1)²+(n+2)²+(n+3)²+(n+4)²=135
n²+n²+2n+1+n²+4n+4+n²+6n+9+n²+8n+16=135
n²+n²+n²+n²+n²+2n+4n+6n+8n=135-1-4-9-16
5n²+20n=105
5n²+20n-105=0

Δ=20²-4*5*(-105)
Δ=400+2100
Δ=2500
√Δ=50

x₁=(-20-50)/(2*5)
x₁=-70/10
x₁=-7

x₂=(-20+50)/(2*5)
x₂=30/10
x₂=3

n=-7 ∨ 3
n+1=-6 ∨ 4
n+2=-5 ∨ 5
n+3=-4 ∨ 6
n+4=-3 ∨ 7

∨ - to znaczy 'lub'

Answer 2

X²+(x+1)²+(x+2)²+(x+3)²+(x+4)²=135
x²+x²+2x+1+x²+4x+4+x²+6x+9+x²+8x+16=135
5x²+20x+30-135=0
5x²+20x-105=0
Δ=b²-4ac
Δ=20²-4×5×(-105)
Δ=400+2100
Δ=2500
√Δ=√2500=50
x₁=(-b-√Δ)/2a=(-20-50)/2×5=-70/10=-7
x₂=(-b+√Δ)/2a=(-20+50)/10=30/10=3

Liczby te to:-7,-6,-5,-4,-3 lub 3,4,5,6,7.

Answer 3

N n+1 n+2 n+3 n+4 <- 5 kolejnych liczb całkowitych

n²+ (n+1)²+(n+2)²+(n+3)²+(n+4)²= 135 n²+n²+2n+1+n²+4n+4+n²+6n+9+n²+8n+16= 135
5n²+20n+30=135
5n²+20n=105
n²+4n=21
n=3
n+1=4
n+2=5
n+3=6
n+4=7