Question

Помогите пожалуйста решить уравнения

Answer 1

5) D=16+40=56

x1= (-4+√56)/2= -2+√14

x2= (-4-√56)/2= -2-√14

6) D= 4+80= 84

x1= (2+√84)/8= (1+√21)/4

x2= (2-√84)/8= (1-√21)/4

7) (8x-3)²=0

x= 3/8

8) D= 144-160= -16

D<0 x∈∅

Answer 2

$displaystyle tt 5) : x^2+4x-10=0\displaystyle tt D=4^2-4cdot1cdot(-10)=16+40=56\displaystyle tt sqrt{D}=sqrt{56}=sqrt{4cdot14}=2sqrt{14}\displaystyle tt x_1=frac{-4+2sqrt{14}}{2}=frac{-2(2-sqrt{14)}}{2}=-(2-sqrt{14})=boxed{bold{-2+sqrt{14}}}\displaystyle tt x_2=frac{-4-2sqrt{14}}{2}=frac{-2(2+sqrt{14})}{2}=-(2+sqrt{14})=boxed{bold{-2-2sqrt{14}}}$

$displaystyle tt 6) : 4x^2-2x-5=0displaystyle tt D=(-2)^2-4cdot4cdot(-5)=4+80=84\displaystyle tt sqrt{D}=sqrt{84}=sqrt{4cdot21}=2sqrt{21}\displaystyle tt x_1=frac{2+2sqrt{21}}{2cdot4}=frac{2(1+sqrt{21})}{8}=boxed{bold{frac{1+sqrt{21}}{4}}}\displaystyle tt x_2=frac{2-2sqrt{21}}{2cdot4}=frac{2(1-sqrt{21})}{8}=boxed{bold{frac{1-sqrt{21}}{4}}}$

$displaystyle tt 7) : 64x^2-48x+9=0\displaystyle tt (8x-3)^2=0\displaystyle tt 8x-3=0\displaystyle tt 8x=3\x=frac{3}{8}\boxed{bold{x=0,375}}$

$displaystyle tt 8) : x^2-12x+40=0\displaystyle tt D=(-12)^2-4cdot1cdot40=144-160=-16\D<0 : to : boxed{bold{x in oslash}}$