Question

determinati valorile parametrului real m pt care ecuatia mx^2-2(m-2)x-m-10=0 are 2 solutii reale de semne contrare

Answer 1

$mx^2~-~2(m-2)x-m-10=0\\ Este~o~ecuatie~de~gradul~al~2-lea,~sub~forma~ax^2+bx+c=0 \\ aici~se~vede~clar~ca~putem~inlocui~m=a \\ -2(m-2)x=-2(a-2)=b \\ -2a+4=b \\ iar~cel~de-al~treilea~factor~c=-m-10~sau~c=-a-10 \\ \Delta=b^2-4ac=(4-2a)^2~+4a(a+10) \\ \Delta=16-16a+4a^2+4a^2+40a \\ \Delta=8a^2+24a+16=8(a+1)(a+2) \\ x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{2a-4+2 \sqrt{2(a+1)(a+2)} }{2a}=\frac{m+ \sqrt{2(m+1)(m+2)}-2 }{m} \\ \\ x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{m- \sqrt{2(m+1)(m+2)}-2}{m}$