Rozwiąż równania: daje naj ;)) a) b).... Question from @Mayakovsky

December 2018 1 13 Report

Rozwiąż równania: daje naj ;))
a) $\frac{ -2x^{2} +x-6 }{3 x^{2} + 4x-4}=0$
b) $\frac{ x^{2} +3x-10 }{ x^{2} + 6x+5}=0$

krysta $a)\\\\ \frac{ -2x^{2} +x-6 }{3 x^{2} + 4x-4}=0\\\\3x^2+4x-4\neq 0\\\\ \Delta =b^2-4ac =4^2 -4*3* (- 4)= 16+48=64 \\ \\\sqrt{\Delta }=\sqrt{64}=87$

$x_{1}=\frac{-b- \sqrt{\Delta } }{2a}=\frac{-4-8}{2 *3}=\frac{-12}{6}=-2 \\ \\x_{2}=\frac{-b+ \sqrt{\Delta } }{2a}=\frac{-4+8}{2 *3}=\frac{4}{6}= \frac{2}{3}\\\\D=R\setminus \left \{ -2,\frac{2}{3} \right \}$

$-2x^{2} +x-6 =0 \\ \\ \Delta =b^2-4ac =1^2 -4*(-2)* (- 6)= 1-48=-47 \\ \\\sqrt{\Delta } <0 \ \ brak \ rozwiazania$

$b)\\\\ \frac{ x^{2} +3x-10 }{ x^{2} + 6x+5}=0\\\\ x^{2} + 6x+5\neq 0\\\\ \Delta =b^2-4ac =6^2 -4*1* 5=36-20=16 \\ \\\sqrt{\Delta }=\sqrt{16}=4$

$x_{1}=\frac{-b- \sqrt{\Delta } }{2a}=\frac{-6-4}{2 }=\frac{-10}{2}=-5\\ \\x_{2}=\frac{-b+ \sqrt{\Delta } }{2a}=\frac{-6+4}{2 }=\frac{-2}{2}=1-\\\\D=R\setminus \left \{-1,-5\right \}$

$x^2 +3x-10 =0 \\ \\ \Delta =b^2-4ac =3^2 -4*1* (-10)=9+40=49\\ \\\sqrt{\Delta }=\sqrt{49}=7 \\ \\x_{1}=\frac{-b- \sqrt{\Delta } }{2a}=\frac{-3-7}{2 }=\frac{-10}{2}=-5 \ \ \not\in D$

$x_{2}=\frac{-b+\sqrt{\Delta } }{2a}=\frac{-3+7}{2 }=\frac{4}{2}=2$

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