Question

Помогите решить уравнения с помощью дисриминанта:

1.(х+4)(в квадрате)=4х(в квадрате) +5

2.9х(4х-1)=3х-1

3.0,09-4х(в квадрате)=1,6х

4.0,1х(в квадрате) -14= -0,4х

5.(х-4)(4х-3)+3=0

6.(х+5)(в квадрате) + (х-2)(в квадрате) + (х-7)(х+7)=11х +80

Answer 1

 

 

 

1)$(x+4)^{2}=4x^{2}+5$

<var />$x^{2}+8x+16=4x^{2}+5$

<var />$-3x^{2}+8x+11=0$

<var> D=64+132=196 </var>

$x_{1}=\frac{8+\sqrt{196}}{6}=\frac{22}{6}=3\frac{4}{6}=3\frac{2}{3}$

<var />$x_{2}=\frac{8-\sqrt{196}}{6}=\frac{-6}{6}=-1$

2)$9x(4x-1)=3x-1$

$36x^{2}-9x-3x+1=0$<var />

$36x^{2}-12x+1=0$<var />

<var>D=144-144=0 </var>

$x_{1}=\frac{12}{2*36}=\frac{12}{72}=\frac{1}{6}$

3)$0,09-4x^{2}=1,6x$

<var />$-4x^{2}-1,6x+0,09=0$

<var> D=2,56+1,44=4</var>

<var />$x_{1}=\frac{1,6+\sqrt{4}}{-8}=\frac{3,6}{-8}=-0,45$

<var />$x_{2}=\frac{1,6-\sqrt{4}}{-8}=\frac{-0,4}{-8}=0,05$

 

 

4) $0,1x^{2}-14=-0,4x$<var />

$0,1x^{2}+0,4x-14=0$<var />

<var>D=0,16+5,6=5,76 </var>

$x_{1}=\frac{-0,4-\sqrt{5,76}}{0,2}=\frac{-0,4-2,4}{0,2}=\frac{-2,8}{0,2}=-1,4$<var />$x_{2}=\frac{-0,4+\sqrt{5,76}}{0,2}=\frac{-0,4+2,4}{0,2}=\frac{2}{0,2}=10$

 

5) $(x-4)(4x-3)+3=0$

$4x^{2}-16x-3x+12+3=0$

$4x^{2}-19x+15=0$

<var> D=361-240=121 </var>

$x_{1}=\frac{19+\sqrt{121}}{8}=\frac{30}{8}=\frac{15}{4}=3\frac{3}{4}$

$x_{2}=\frac{19-\sqrt{121}}{8}=\frac{8}{8}=1$

6)$(x+5)^{2}+(x-2)^{2}+(x-7)(x+7)=11x+80$

$x^{2}+10x+25+x^{2}-4x+4+x^{2}-49+11x-80=0$<var />

$3x^{2}-5x-100=0$<var />

<var>D=25+1200=1225 </var>

$x_{1}=\frac{5+\sqrt{1225}}{6}=\frac{40}{6}=\frac{20}{3}=6\frac{2}{3}$<var />

$x_{2}=\frac{5-\sqrt{1225}}{6}=\frac{-30}{6}=-6$