Question

Помогите срочно нужно
Решите уравнения:
1). X^2-3x-4=0
2). 4x^2+4x+1=0
3). 5^2x =25
4). 3^5x+1 = 27
5). 3x^4+5x^2-8=0

Answer 1

$x^2-3x-4=0 \ D=b^2-4ac=(-3)^2-4*1*(-4)=9+16=25 \ x_{1} = frac{-b+ sqrt{D} }{2a} = frac{3+5}{2} = frac{8}{2} =4 \ \ x_{2} = frac{-b- sqrt{D} }{2a} = frac{3-5}{2} = frac{-2}{2} =-1 \ \ (x-4)(x+1)=0$


$4x^2+4x+1=0 \ D=b^2-4ac=4^2-4*4*1=16-16=0 \ x_{1,2} = frac{-b}{2a} = frac{-4}{8} =- frac{1}{2} \ \ (2x+1)(2x+1)=0$


$5^{2x} =25 \ 5^{2x} =5^2 \ 2x=2 \ x=1$


$3^{5x+1} = 27 \ 3^{5x+1} =3^3 \5x+1=3 \ 5x=2 \ x= frac{2}{5}$


$3x^4+5x^2-8=0 \ x^2=t \ 3t^2+5t-8=0 \ D=b^2-4ac=5^2-4*3*(-8)=25+96=121 \ t_{1} = frac{-b+ sqrt{D} }{2a} = frac{-5+11}{6} = frac{6}{6} =1 \ \ t_{2} = frac{-b- sqrt{D} }{2a} = frac{-5-11}{6} = frac{-16}{6} =- frac{8}{3} =-2 frac{2}{3} \ \ x^2=1 \ x_{1} =1,x_{2} =-1$