Question

ex5...................​

Answer 1

a) u=ma+(2m-1)b

v=(3m-2)a+mb

Pentru ca u si v sa fie coliniari trebuie indeplinita conditia

$\frac{m}{3m-2}=\frac{2m-1}{m}\\\\m*m=(3m-2)*(2m-1)\\\\m^2=6m^2-3m-4m+2\\\\m^2=6m^2-7m+2\\\\5m^2-7m+2=0\\\\\Delta=b^2-4ac\\\\\Delta=49-40=9\\\\m_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{7+3}{10}=1\\\\m_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{7-3}{10}=\frac{4}{10}=0.4$

b)u=(m²+1)a+ (2m-1)b

v=(5+m)a+(m-5)b

$\frac{m^2+1}{5+m}=\frac{2m-1}{m-5}\\\\(m^2+1)*(m-5)=(m+5)*(2m-1)\\\\m^3-5m^2+m-5=2m^2-m+10m-5\\\\m^3-5m^2+m-5-2m^2+m-10m+5=0\\\\m^3-7m^2-8m=0\\\\m*(m^2-7m-8)=0\\\\m_1=0\\\\m_2^2-7m_2-8=0\\\\\Delta=49+32=81\\\\m_{21}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{7+9}{2}=\frac{16}{2}=8\\\\m_{22}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{7-9}{2}=\frac{-2}{2}=-1$