2 szkoły ponadpodstawowej

II liceum

II technikum

Matematyka 2.  Zakres podstawowy. Po gimnazjum

Rozwiązanie 1

Będziemy korzystać z tego, że:  p=−2ab​   q=−4aΔ​     a) p=0,   q=−1,   a=4      −1=−4⋅4Δ​   ⇒   −16=−Δ   ⇒   Δ=16&gt;0      Δ<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
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M834 80h400000v40h-400000z'/></svg>​=16<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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Δ=(−6)2−4⋅1⋅9=36−36=0

f(x)=−(x−2)(x+2)=−(x2−22)=−(x2−4)=−x2+4

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