[tex]zad.a\\
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y=5x^{2} -3x\\
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a=5,~~b=-3,~~c=0\\
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\Delta=(-3)^{2} -4\cdot 5\cdot 0=9\\
\sqrt{\Delta} =\sqrt{9} =3\\
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x_{1} =\dfrac{3-3}{10}=0 ~~\lor~~x_{2} =\dfrac{3+3}{10} =\dfrac{3}{5} \\
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y=5x\cdot (x-\dfrac{3}{5})~~postac~~iloczynowa[/tex]
[tex]zad.b\\
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y=16x^{2} -1\\
a=16,~~b=0,~~c=-1\\
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\Delta=0^{2} -4\cdot 16\cdot (-1)=64\\
\sqrt{\Delta} =\sqrt{64} =8\\
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x_{1} =\dfrac{0-8}{32} =-\frac{1}{4} ~~\lor~~x_{2} =\dfrac{0+8}{32} =\frac{1}{4} \\
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y=16\cdot (x-\dfrac{1}{4} )\cdot (x+\dfrac{1}{4} )~~postac~~iloczynowa[/tex]
[tex]zad.c\\
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y=9x^{2} -8\\
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a=9,~~b=0,~~c=-8\\
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\Delta = 0^{2} -4\cdot 9\cdot (-8)=288\\
\sqrt{\Delta} =\sqrt{288} =12\sqrt{2} \\
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x_{1} =\dfrac{0-12\sqrt{2} }{18} =-\dfrac{2\sqrt{2} }{3} ~~\lor~~x_{2} =\dfrac{0+12\sqrt{2} }{18} =\dfrac{2\sqrt{2} }{3} \\
\\
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y=9(x-\dfrac{2\sqrt{2} }{3})\cdot (x + \dfrac{2\sqrt{2} }{3})~~postac~~ iloczynowa[/tex]
Korzystam :
[tex]f(x)=ax^{2} +bx+c~~funkcja~~kwadratowa~~w~~postaci~~ogolnej\\
\Delta~~-~~trojmian~~kwadratowy~~nazywany~~delta\\
\Delta=b^{2} -4\cdot a\cdot c\\
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x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\
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f(x)=a\cdot (x-x_{1} )\cdot (x -x_{2} )~~postac~~iloczynowa ~~funkcji~~kwadratowej[/tex]
