3.29-суретте штрихталған үшбұрыштың периметрі p(x)-ті ed=x-ке тәуелді формуламен анықтаңдар. Мұнда AB=8, BC=6, AC=10 және ED||BC​

2x² -3x -9 =0 ;
2x² -6x +3x -9 =0;
2x(x-3) +3(x-3) =0 ;
(x-3)(2x+3) =0 ;
2(x+3/2)(x-3) =0 ⇒[ x+3/2=0 ;x-3 =0 ⇔[  x= - 3/2 ;x=3.
* * *x₁ =-3/2 ; x₂ =3.* * *

5 x²+3х+4=5(x² +(3/5)x +4/5 ) =5(x² +2*(3/10)x +(3/10)² -(3/10)² +4/5 ) =
=5 ((x +3/10)² +71/100) ≠0 →не имеет действительных корней.

3x² -2x-8 =3(x² -2*(1/3)*x +(1/3)² - (1/3)² -8/3) =3*((x-1/3)² -(5/3)²) =
3*(x-1/3 -5/3)(x-1/3 +5/3) =3*(x-2)(x +4/3) .
3x² -2x-8 =0 ⇔3*(x-2)(x +4/3) =0 x⇔[ x-2 =0 ;x +4/3=0⇔[ x=2 ; x = - 4/3.
* * *
3x² -2x-8 =(3x² -6x) +(4x -8) =3x(x-2) +4(x-2) =(x-2)(3x+4) =3(x+4/3)(x-2) .
3x² -2x-8 =0 ⇔3(x+4/3)(x-2) =0 ⇒ [x+4/3=0 ;x-2 =0 ⇔[x= - 4/3 ;x=2 .
* * * x₁ = - 4/3 ;x₂ =2 . * * *

2x² -3x -9 =0 ;
2x² -6x +3x -9 =0;
2x(x-3) +3(x-3) =0 ;
(x-3)(2x+3) =0 ;
2(x+3/2)(x-3) =0 ⇒[ x+3/2=0 ;x-3 =0 ⇔[  x= - 3/2 ;x=3.
* * *x₁ =-3/2 ; x₂ =3.* * *

5 x²+3х+4=5(x² +(3/5)x +4/5 ) =5(x² +2*(3/10)x +(3/10)² -(3/10)² +4/5 ) =
=5 ((x +3/10)² +71/100) ≠0 →не имеет действительных корней.

3x² -2x-8 =3(x² -2*(1/3)*x +(1/3)² - (1/3)² -8/3) =3*((x-1/3)² -(5/3)²) =
3*(x-1/3 -5/3)(x-1/3 +5/3) =3*(x-2)(x +4/3) .
3x² -2x-8 =0 ⇔3*(x-2)(x +4/3) =0 x⇔[ x-2 =0 ;x +4/3=0⇔[ x=2 ; x = - 4/3.
* * *
3x² -2x-8 =(3x² -6x) +(4x -8) =3x(x-2) +4(x-2) =(x-2)(3x+4) =3(x+4/3)(x-2) .
3x² -2x-8 =0 ⇔3(x+4/3)(x-2) =0 ⇒ [x+4/3=0 ;x-2 =0 ⇔[x= - 4/3 ;x=2 .
* * * x₁ = - 4/3 ;x₂ =2 . * * *