Question

помогите решить через дискриминант​

Answer 1

$\displaystyle x^2 -10x + 24 = 0 \\\\ a= \bold1 ~ ; ~ b =\bold{-10} ~ ; ~ c = \bold{24} \\\\ D= \boldsymbol{b^2 -} 4 \boldsymbol{ac = 100 - 96 = 4}\\\\\\ x_1= \cfrac{-b-\sqrt{D} }{2a} =\boldsymbol{ \frac{10 -2}{2\cdot 1} = 4} \\\\\\ x_2 = \cfrac{-b +\sqrt{D} }{2a} = \boldsymbol{\frac{10+2}{2\cdot 1} = 6 }\\\\\\ x^2 -10x + 24 = \\\\ = (x-\boldsymbol{x_1})(x-\boldsymbol{x_2}) = \boldsymbol{(x-4)(x-6)}$          

                                                         

$\displaystyle 4x^2 -11x -3 = 0 \\\\ a= \bold4 ~ ; ~ b =\bold{-11} ~ ; ~ c = \bold{-3} \\\\ D= \boldsymbol{b^2 -} 4 \boldsymbol{ac = 121 + 48 = 169 = 13^2 }\\\\\\ x_1= \cfrac{-b-\sqrt{D} }{2a} =\boldsymbol{ \frac{11 -13}{2\cdot 4} = - \frac{1}{4} } \\\\\\ x_2 = \cfrac{-b +\sqrt{D} }{2a} = \boldsymbol{\frac{11+13}{2\cdot 4} = 3 }\\\\\\ 4x^2 -11x -3 = \\\\ = (x-\boldsymbol{x_1})(x-\boldsymbol{x_2}) = \boldsymbol{ \bigg(x+ \frac{1}{4} \bigg )(x-3)}$

                                                         

$7x^2 - 8x + 1= 0 \\\\\\ D = 64 -28 = 36 \\\\\\ x_1= \cfrac{8 - 6}{14} = - \cfrac{1}{7 } \\\\\\ x_2= \cfrac{8+ 6}{14} = 1 \\\\\\ 7x^2 - 8x +1 = \\\\\\ \bigg (x + \cfrac{1}{7} \bigg )( x -1) =0$