Question

ПОМОГИТЕ ПОЖАЛУЙСТА С АЛГЕБРОЙ!!!!!!!!!!!Нужно решить уравнения:1) х²+(√х)²-6=02) х²+3(√х)²-4=03) √2х²+4√3х-2√2=04) у²+2(√3+1)у+2√3=0

Answer 1

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

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

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

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