a)  $y=-x^2-2mx+m-1$  

$\Delta ={\left(-2m\right)}^2-4\cdot \left(-1\right)\cdot \left(m-1\right)=4m^2+4m-4$  

$q=1$  

$\frac{-\Delta }{4a}=1$  

$\frac{-\left(4m^2+4m-4\right)}{4\cdot \left(-1\right)}=1$  

$\frac{-4\left(m^2+m-1\right)}{-4}=1$  

$m^2+m-1=1\mid -1$  

$m^2+m-2=0$  

${\Delta }_m=1^2-4\cdot 1\cdot \left(-2\right)=1+8=9$  

$\sqrt{{\Delta }_m}=3$  

$m_1=\frac{-1-3}{2}=\frac{-4}{2}=-2$  

$m_2=\frac{-1+3}{2}=\frac{2}{2}=1$  

 

Odpowiedź:  $m\in \left\{-2,1\right\}$  

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b)  $y=\left(m-1\right)x^2+\left(m-1\right)x+m+1$  

Założenie:  $m-1\ne 0\text{czyli}m\ne 1$  

$\Delta ={\left(m-1\right)}^2-4\cdot \left(m-1\right)\cdot \left(m+1\right)=m^2-2m+1-4\left(m^2-1\right)=m^2-2m+1-4m^2+4=-3m^2-2m+5$  

$q=1$  

$\frac{-\Delta }{4a}=1$  

$\frac{-\left(-3m^2-2m+5\right)}{4\left(m-1\right)}=1\mid \cdot 4\left(m-1\right)$  

$\left(3m^2+2m-5\right)=4\left(m-1\right)$  

$3m^2+2m-5-4m+4=0$  

$3m^2-2m-1=0$  

${\Delta }_m={\left(-2\right)}^2-4\cdot 3\cdot \left(-1\right)=4+12=16$  

$\sqrt{{\Delta }_m}=4$  

$m_1=\frac{2-4}{2\cdot 3}=\frac{-2}{6}=-\frac{1}{3}$  

$m_2=\frac{2+4}{2\cdot 3}=\frac{6}{6}=1\text{sprzecznosˊcˊ}$  

 

Odpowiedź:  $m=-\frac{1}{3}$