Question

ПОМОГИТЕ ПОЖАЛУЙСТА С АЛГЕБРОЙ!!!!!!!!!!!
Нужно решить уравнения:
1) х²+(√х)²-6=0
2) х²+3(√х)²-4=0
3) √2х²+4√3х-2√2=0
4) у²+2(√3+1)у+2√3=0

Answer 1

$1) x^{2} +( sqrt{x} ) ^{2} -6=0 \ x^{2} +|x|-6=0 \ \ x^{2} +x-6=0 \ D=1+24=25 \ sqrt{D} =5 \ x_{1} = frac{-1+5}{2} = frac{4}{2} =2 \ x_{2} = frac{-1-5}{2} =-3 \ ili \ x^{2} -x-6=0 \ D=1+24=25 \ sqrt{D} =5 \ x_{1} = frac{1+5}{2} =3 \ x_{2} = frac{1-5}{2} =-2$

$2) x^{2} +3( sqrt{x} ) ^{2} -4=0 \ x^{2} +3|x|-4=0 \ \ x^{2} +3x-4=0 \ D=9+16=25 \ sqrt{D} =5 \ x_{1} = frac{-3+5}{2} =1 \ x_{2}= frac{-3-5}{2} =-4 \ \ x^{2} -3x-4=0 \ D=9+16=25 \ sqrt{D} =5 \ x_{1} = frac{3+5}{2} =4 \ x_{2} = frac{3-5}{2} =-1$

$3) sqrt{2} x^{2} +4 sqrt{3} x-2 sqrt{2} =0 \ D=(4 sqrt{3} ) ^{2} -4* sqrt{2} *(-2 sqrt{2} )=48+4*2*2=48+16=64 \ sqrt{D} =8 \ x_{1} = frac{-4 sqrt{3} +8}{2 sqrt{2} } = frac{2(-2 sqrt{3} +4)}{ 2sqrt{2} } = frac{-2 sqrt{3} +4}{ sqrt{2} } \ x_{2} = frac{-4 sqrt{3} -8}{2 sqrt{2} } = frac{2(-2 sqrt{3} -4)}{ 2sqrt{2} } = frac{-2 sqrt{3} -4}{ sqrt{2} }$

$4) y ^{2} +2( sqrt{3} +1)y+2 sqrt{3} =0 \ y^{2} +(2 sqrt{3} +2)y+2 sqrt{3} =0 \ D=(2 sqrt{3} +2) ^{2} -4*2 sqrt{3} =12+4+8 sqrt{3} -8 sqrt{3} =16 \ sqrt{D} =4 \ y _{1} = frac{-2 sqrt{3} -2-4}{2} = frac{-2 sqrt{3}-6 }{2} =- sqrt{3} -3 \ y _{2} = frac{-2 sqrt{3} -2+4}{2} = frac{-2 sqrt{3}+2 }{2} = sqrt{3} +1$