Question

(3x-2)²-7x(x-3)<0 помогите

Answer 1

Ответ:

x ∈ (-4; -0,5)

Объяснение:

$(3x-2)^2-7x(x-3) < 0\\9x^2-12x+4-7x^2+21x < 0\\2x^2+9x+4 < 0\\2x^2+9x+4=0\\D=b^2-4ac=9^2-4\cdot2\cdot4=81-32=49\\\displaystyle x_1=\frac{-b-\sqrt{D} }{2a} =\frac{-9-\sqrt{49} }{2\cdot2} =\frac{-9-7}{4} =-16:4=-4\\x_2=\frac{-b+\sqrt{D} }{2a} =\frac{-9+\sqrt{49} }{2\cdot2}=\frac{-9+7}{4} =-2:4=-0,5$

$\Large \boldsymbol {} x\in \displaystyle\left(-4;~-0,5\right).$