Oblicz pierwiastki trójmianu kwadratowego. a) y=x^2 + x - 6 b) y+x^2 + 10x + 21 c) y=x^2

December 2018 1 10 Report

Oblicz pierwiastki trójmianu kwadratowego.
a) y=x^2 + x - 6
b) y+x^2 + 10x + 21
c) y=x^2 - x +1/4
d) y=2x^2 + 9x - 5
e) y=2x^2 + 4x - 1
f) y=3x^2 + x - 2
G) y=-2x^2 + 2x + 1
h) y=-6x^2 - 7x +3
i) y=-3x^2 + 5x - 4
PROSZĘ POTRZEBNE NA JUTRO!!!

ag94 A) $y=ax^2+bx+c\\
y=x^2 + x - 6\\
\Delta=b^2-4ac=1^2-4\cdot1\cdot(-6)=1+24=25\\
 \sqrt{\Delta} = \sqrt{25} = \sqrt{5^2}=5 \\\\
x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-1- 5}{2\cdot1} =\frac{-6}{2}=-3 \\\\ 
x_2= \frac{-b+ \sqrt{\Delta}}{2a} =\frac{-1+ 5}{2\cdot1} =\frac{4}{2}=2$

b) $y=x^2 + 10x + 21\\\Delta=b^2-4ac=10^2-4\cdot1\cdot21=100-84=16\\ \sqrt{\Delta} = \sqrt{16} = \sqrt{4^2}=4 \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-10- 4}{2\cdot1} =\frac{-14}{2}=-7 \\\\ x_2= \frac{-b+ \sqrt{\Delta}}{2a} =\frac{-10+ 4}{2\cdot1} =\frac{-6}{2}=-3$

c) $y=x^2 - x + \frac{1}{4} \\\\ \Delta=b^2-4ac=(-1)^2-4\cdot1\cdot \frac{1}{4} =1-1=0\\ \\x_0= \frac{-b}{2a} =\frac{-(-1)}{2\cdot1} =\frac{1}{2}= \frac{1}{2}$

d)
$y=2x^2 + 9x - 5 \\\Delta=b^2-4ac=9^2-4\cdot2\cdot(-5) =81+40=121\\ \sqrt{\Delta} = \sqrt{121} = \sqrt{11^2}=11 \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-9- 11}{2\cdot2} =\frac{-20}{4}=-5 \\\\ x_2= \frac{-b+ \sqrt{\Delta}}{2a} =\frac{-9+ 11}{2\cdot2} =\frac{2}{4}= \frac{1}{2}$

e)
$y=2x^2 + 4x - 1 \\\Delta=b^2-4ac=4^2-4\cdot2\cdot (-1) =16+8=24\\ \sqrt{\Delta} = \sqrt{24} = \sqrt{2^2\cdot6}=2 \sqrt{6} \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-4- 2 \sqrt{6} }{2\cdot2} =\frac{-2- \sqrt{6} }{2}=- \frac{1}{2}(-2- \sqrt{6}) \\\\ x_2= \frac{-b+ \sqrt{\Delta}}{2a} =\frac{-4+ 2 \sqrt{6} }{2\cdot2} =\frac{-2+ \sqrt{6} }{2}=- \frac{1}{2}(-2+\sqrt{6})$

f)
$y=3x^2 + x - 2 \\\Delta=b^2-4ac=1^2-4\cdot3\cdot (-2) =1+24=25\\ \sqrt{\Delta} = \sqrt{25} = \sqrt{5^2}=5 \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-1-5}{2\cdot3} =\frac{-6}{6}=-1 \\\\ x_2= \frac{-b+ \sqrt{\Delta}}{2a} =\frac{-1+5}{2\cdot3} =\frac{4}{6}= \frac{2}{3}$

G)
$y=-2x^2 + 2x + 1 \\\Delta=b^2-4ac=2^2-4\cdot(-2)\cdot 1 =4+8=12\\ \sqrt{\Delta} = \sqrt{12} = \sqrt{2^2\cdot3}=2 \sqrt{3} \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-2- 2 \sqrt{3}}{2\cdot(-2)} =\frac{1+ \sqrt{3} }{2}= \frac{1}{2}(1+ \sqrt{3} ) \\\\x_2= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-2+ 2 \sqrt{3}}{2\cdot(-2)} =\frac{1- \sqrt{3} }{2}= \frac{1}{2}(1- \sqrt{3} )$

h)
$y=-6x^2 - 7x +3 \\\Delta=b^2-4ac=(-7)^2-4\cdot(-6)\cdot 3 =49+72=121\\ \sqrt{\Delta} = \sqrt{121} = \sqrt{11^2}=11 \\\\x_1= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-(-7)- 11}{2\cdot(-6)} =\frac{-4}{-12}= \frac{1}{3} \\\\x_2= \frac{-b- \sqrt{\Delta}}{2a} =\frac{-(-7)+ 11}{2\cdot(-6)} =\frac{18}{-12}= -\frac{3}{2}$

i)
$y=-3x^2 + 5x - 4 \\\Delta=b^2-4ac=5^2-4\cdot(-3)\cdot (-4) =25-48=-23<0 \\
brak\ rozwiazan$

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