Question

1) 5x2-12x+7=0 2) 3y2 +7y+4=0​

Answer 1

Ответ:

1) 5x2-12x+7=0

$5x^{2} -12x+7=0;$

Используем формулу квадратного уравнения $ax^{2} +bx+c=0$  и решим с помощью дискриминанта D:

$a=5; b=-12;c=7;\\\\D=b^{2} -4*a*c=(-12)^{2} -4*5*7=144-140=4;\\\\\sqrt{D}=\sqrt{4} =2.\\\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-(-12)+2}{2*5} =\frac{14}{10} =\frac{7}{5} ;\\\\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-(-12)-2}{2*5} =\frac{12-2}{10} =\frac{10}{10}=1.$

$x_{1} =\frac{7}{5}$    или $x_{1} =1\frac{2}{5};$

$x_{2}=1.$

2) 3y2 +7y+4=0​

$3y^{2} +7y+4=0;\\\\a=3; b=7;c=4;\\\\D=b^{2} -4*a*c=7^{2} -4*3*4=49-48=1;\\\\\sqrt{D}=\sqrt{1} =1.\\\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-7+1}{2*3} =\frac{-6}{6} =-1 ;\\\\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-7-1}{2*3} =\frac{-8}{6} =-\frac{4}{3}.$

$x_{1} =-1;\\\\x_{2}=-\frac{4}{3}=-1\frac{1}{3}.$