Question

daca 21 supra x la puterea a 2 +4x =7/4, atunci:

Answer 1

$\frac{21}{x^2+4x}=\frac{7}{4}\\21\cdot \:4=\left(x^2+4x\right)\cdot \:7\\\\84=\left(x^2+4x\right)\cdot \:7\\84=7x^2+7\cdot \:4x\\84=7x^2+28x\\7x^2+28x-84=0\\\\a=7,\:b=28,\:c=-84:\quad x_{1,\:2}=\frac{-28\pm \sqrt{28^2-4\cdot \:7\left(-84\right)}}{2\cdot \:7}\\x_1=2,\:x_2=-6\\\\D.  x<3$