Matematică, întrebare adresată de albert69, 8 ani în urmă

va rog am nevoie cat mai rapid, dau 30 puncte , ma abonez.. de toate

x²-9x+14=0
5x²-4x+1=0
3x²-4x+1=0
x²-11×+30=0​


albert69: sunt ecuatii de gradul 2
Excelsis: da

Răspunsuri la întrebare

Răspuns de tcostel
13

 

\displaystyle\bf\\a)\\x^2-9x+14=0\\x^2-2x-7x+14=0\\x(x-2)-7(x-2)=0\\(x-2)(x-7)=0\\x-2=0\implies x_1=2\\x-7=0\implies x_2=7\\\\b)\\5x^2-4x+1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-20}}{10} =\frac{4\pm\sqrt{-4}}{10}=\frac{4\pm2i}{10}=\frac{2\pm i}{5} \\\\x_{12}\notin R.\\Dar:\\5x^2-4x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16+20}}{10} =\frac{4\pm\sqrt{36}}{10}=\frac{4\pm6}{10}=\frac{2\pm 3}{5}\\\\x_1=\frac{2-3}{5}=-\frac{1}{5}\\x_2=\frac{2+3}{5}=1

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\displaystyle\bf\\c)\\3x^2-4x+1=0\\\\x{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-12}}{6}=\frac{4\pm\sqrt{4}}{6}=\frac{4\pm2}{6}=\frac{2\pm1}{3}\\\\x_1=\frac{2-1}{3}=\frac{1}{3}\\\\x_1=\frac{2+1}{3}=\frac{3}{3}=1\\\\d)\\x^2-11x+30=0\\x^2-5x-6x+30=0\\x(x-5)-6(x-5)=0\\\\(x-5)(x-6)=0\\x-5=0\implies~x_1=5\\x-6=0\implies~x_2=6

 

 

 


nicadarc2013: gg
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