Odpowiedź :
[tex]a)\\x^{2}+10x + 25 = 0\\\\(x+5)^{2} = 0\\\\x+5 = 0\\\\x_{o} = -5[/tex]
[tex]b)\\x^{2}+4x-21 = 0\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot1\cdot(-21) = 16 + 84 = 100\\\\\sqrt{\Delta} = \sqrt{100} = 10\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-4-10}{2} = -7\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} =\frac{-4+10}{2} = 3[/tex]
[tex]c)\\4x^{2}-8x = 0 \ \ /:4\\\\x^{2}-2x = 0\\\\x(x - 2) = 0\\\\x = 0 \ \vee \ x-2 = 0\\\\x = 0 \ \vee \ x = 2[/tex]
Odpowiedź:
a) x²+10x+25=0
∆=b²-4ac=10²-4×1×25=100-100=0
∆=0, więc równanie ma tylko jedno rozwiązanie
x=-b/2a = -10/2×1 = -10/2 =-5
Odp. x=-5
b) x²+4x-21=0
∆=b²-4ac= 4²-4×1×(-21) =16+84=100
√∆=√100=10
∆>0, więc ma dwa rozwiązania
x1= -b-√∆ |2a = -4-10|2×1= -14/2=-7
x2= -b+√∆ |2a= -4+10|2=6|2=3
Odp. x=-7 v x=3
c) 4x²-8x=0
x(4x-8) =0
x=0
4x-8=0
4x=8/:4
x=2
Odp. x=0 v x=2