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

$\displaystyle \tt 5) \: x^2+4x-10=0\\\displaystyle \tt D=4^2-4\cdot1\cdot(-10)=16+40=56\\\displaystyle \tt \sqrt{D}=\sqrt{56}=\sqrt{4\cdot14}=2\sqrt{14}\\\displaystyle \tt x_1=\frac{-4+2\sqrt{14}}{2}=\frac{-2(2-\sqrt{14)}}{2}=-(2-\sqrt{14})=\boxed{\bold{-2+\sqrt{14}}}\\\displaystyle \tt x_2=\frac{-4-2\sqrt{14}}{2}=\frac{-2(2+\sqrt{14})}{2}=-(2+\sqrt{14})=\boxed{\bold{-2-2\sqrt{14}}}$

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

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

$\displaystyle \tt 8) \: x^2-12x+40=0\\\displaystyle \tt D=(-12)^2-4\cdot1\cdot40=144-160=-16\\D<0 \: \to \: \boxed{\bold{x \in \oslash}}$