Question

1) 2x^2-7x+6>0
2) -5x^2+11x-6>0
Помогите пожалуйста решить​

Answer 1

( 1 )

$2x^2-7x+6>0\\\\2x^2-7x+6=0\\\\D=(-7)^2-4*2*6=49-48=1\\\\x_1=\frac{-b+\sqrt{D} }{2a} =\frac{7+\sqrt{1} }{2*2} =\frac{7+1}{4}=\frac{8}{4}=2\\\\x_2=\frac{-b-\sqrt{D} }{2a} =\frac{7-\sqrt{1} }{2*2} =\frac{7-1}{4}=\frac{6}{4}=1\frac{2}{4}= 1\frac{1}{2} =1,5$

x ∈ ( -∞ ; 1,5 ) ∪ ( 2 ; +∞ )

( 2 )

$-5x^2+11x-6>0\\\\-5x^2+11x-6=0\\\\D=11^2-4*(-5)*(-6)=121-120=1\\\\x_1=\frac{-b+\sqrt{D} }{2a} =\frac{-11+\sqrt{1} }{2*(-5)} =\frac{-11+1}{-10}=\frac{-10}{-10}=1\\\\x_2=\frac{-b-\sqrt{D} }{2a} =\frac{-11-\sqrt{1} }{2*(-5)} =\frac{-11-1}{-10}=\frac{-12}{-10}=1\frac{-2}{-10}= 1,2$

x ∈ ( 1  ; 1,2 )