Question

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Answer 1

$\displaystyle\bf\\1)\\\\x^{2} +2mx+m^{2} =0\\\\(x+m)^{2} =0\\\\x_{1}=-2\\\\(-2+m)^{2} =0\\\\m-2=0\\\\\boxed{m=2}\\\\\\2)\\\\m^{2}x^{2} -4 +9mx-6=0\\\\m^{2} x^{2} +9mx-10=0\\\\x_{1} =2\\\\m^{2} \cdot 2^{2} +9m\cdot 2-10=0\\\\4m^{2} +18m-10=0\\\\D=18^{2} -4\cdot 4\cdot (-10)=324+160=484=22^{2} \\\\\\m_{1} =\frac{-18-22}{8} =-5\\\\\\m_{2} =\frac{-18+22}{8} =0,5\\\\\\\boxed{m_{1} =-5 \ \ ; \ \ m_{2} =0,5}$