proszę o rozwiązanie równania 13^2= x^2 + (x+7)^2
13²=x²+(x+7)²
x²+x²+49+14x=169
2x²+14x-120=0
x²+7x-60=0
a=1
b=7
c=-60
Δ=b²-4ac
Δ=7²-4·1·(-60)=49+240=289
√Δ=17
x₁=17-7/2=5
x₂=-17-7/2=-12
13^2 = x^2 + (x+7)^2
x^2 + (x+7)^2 - 13^2 = 0
x^2 + x^2 + 14x + 49 - 169 = 0
2x^2 + 14x - 120 = 0 /:2
x^2 + 7x - 60 = 0
a = 1, b - 7, c = -60
D(delta) = b^2 - 4ac
D = 49-4*1*(-60) = 49+240 = 289
VD = 17
x1 = (-b-VD)/2a = (-7-17)/2 = -24/2 = -12
x2 = (-b+VD)/2a = (-7+17)/2 = 10/2 = 5
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13²=x²+(x+7)²
x²+x²+49+14x=169
2x²+14x-120=0
x²+7x-60=0
a=1
b=7
c=-60
Δ=b²-4ac
Δ=7²-4·1·(-60)=49+240=289
√Δ=17
x₁=17-7/2=5
x₂=-17-7/2=-12
13^2 = x^2 + (x+7)^2
x^2 + (x+7)^2 - 13^2 = 0
x^2 + x^2 + 14x + 49 - 169 = 0
2x^2 + 14x - 120 = 0 /:2
x^2 + 7x - 60 = 0
a = 1, b - 7, c = -60
D(delta) = b^2 - 4ac
D = 49-4*1*(-60) = 49+240 = 289
VD = 17
x1 = (-b-VD)/2a = (-7-17)/2 = -24/2 = -12
x2 = (-b+VD)/2a = (-7+17)/2 = 10/2 = 5