Question

sa se determine parametrul real m astfel incat ecuatia mx^2-2(m+1)x+m-5=0 sa nu aiba solutii reale

Answer 1

O ecuatie de gradul doi nu are solutii reale daca Δ < 0.
Δ = $b^{2}-4ac$
Iar in aceasta ecuatie:
$mx^{2}-2(m+1)x+m-5=0$
b = -2(m+1)
a = m
c = m-5
Deci, ecuatia nu are solutii reale daca
$(-2(m+1))^{2}-4m(m-5)\ \textless \ 0$                                      $4(m+1)^{2}-4m(m-5) \ \textless \ 0$
$4m^{2}+8m+4-4m^{2}+20m\ \textless \ 0$
$28m+4\ \textless \ 0$
$28m\ \textless \ -4$
$m\ \textless \ -\frac{4}{28}$
$m\ \textless \ -\frac{1}{7}$