Question

Разложите на множители квадратный трехчлен: 1) 0,8x²-19,8x-5 2) 3,5- + 3) x²+x√2-2 4) x²-x√6+1

Answer 1

$0.8x^2-19,8x-5=0 \\\ D_1=9.9^2-0.8\cdot(-5)=102.01=10.1^2 \\\ x_1= \frac{9.9+10.1}{0.8} =25 \\\ x_2= \frac{9.9-10.1}{0.8} = -\frac{1}{4} \\\ 0.8x^2-19,8x-5=0.8(x-25)(x+ \frac{1}{4} )=(x-25)(0.8x+ 0.2 )$

$\frac{2}{3}x^2-3 \frac{1}{3}x+3.5=0 \\\ \frac{2}{3}x^2- \frac{10}{3}x+ \frac{7}{2} =0 \\\ D_1=(- \frac{5}{3} )^2- \frac{2}{3} \cdot \frac{7}{2} =\frac{25}{9} - \frac{21}{9}= \frac{4}{9} =( \frac{2}{3} )^2 \\\ x_1= \cfrac{ \frac{5}{3}+ \frac{2}{3} }{ \frac{2}{3} } = \frac{ 7 }{2 } \\\ x_2= \cfrac{ \frac{5}{3}- \frac{2}{3} }{ \frac{2}{3} } = \frac{ 3 }{2 } \\\ \frac{2}{3}x^2-3 \frac{1}{3}x+3.5= \frac{2}{3}(x- \frac{7}{2} )( x- \frac{3}{2} )= (x- \frac{7}{2} )( \frac{2}{3}x- 1 )$

$x^2+x \sqrt{2} -2=0 \\\ D=( \sqrt{2} )^2-4\cdot1\cdot(-2)=2+8=10 \\\ x_1= \frac{- \sqrt{2} - \sqrt{10} }{2} \\\ x_2= \frac{- \sqrt{2} + \sqrt{10} }{2} \\\ x^2+x \sqrt{2} -2=(x-\frac{- \sqrt{2} - \sqrt{10} }{2} )(x-\frac{- \sqrt{2} + \sqrt{10} }{2} ) \\\ x^2+x \sqrt{2} -2=(x+\frac{ \sqrt{2} + \sqrt{10} }{2} )(x+\frac{ \sqrt{2} - \sqrt{10} }{2} )$

$x^2-x \sqrt{6}+1 =0 \\\ D=(- \sqrt{6} )^2-4\cdot1\cdot1=6-4=2 \\\ x_1= \frac{ \sqrt{6}+ \sqrt{2} }{2} \\\ x_2= \frac{ \sqrt{6}- \sqrt{2} }{2} \\\ x^2-x \sqrt{6}+1 =(x-\frac{ \sqrt{6}+ \sqrt{2} }{2} )(x-\frac{ \sqrt{6}- \sqrt{2} }{2} )$