Question

Розв'яжіть рівняння, застосовуючи теорему
Вієта:
1)х^2+17х+60=0
2)х^2-7х+10=0
3)х^2+3х-18=0
4) х^2-5х-24=0
5)х^2+9х-22=0
6)х^2+6х+8=0

Answer 1

Ответ:

1)$x_1=-12, x_2=-5$

2)$x_1=2, x_2=5$

3)$x_1=-6, x_2=3$

4)$x_1=-3, x_2=8$

5)$x_1=-11, x_2=2$

6)$x_1=-4, x_2=-2$

Объяснение:

1) $x^{2} +17x+60=0\\x^{2} +12x+5x+60=0\\x*(x+12)+5*(x+12)=0\\(x+12)*(x+5)=0\\\left \{ {{x+12=0} \atop {x+5=0}} \right. \\\left \{ {{x=-12} \atop {x=-5}} \right. \\x_1=-12, x_2=-5$2) $x^{2} -7x+10=0\\x^{2} -2x-5x+10=0\\x*(x-2)-5*(x-2)=0\\(x-2)*(x-5)=0\\\left \{ {{x-2=0} \atop {x-5=0}} \right. \\\left \{ {{x=2} \atop {x=5}} \right. \\x_1=2, x_2=5$

3)$x^{2} +3x-18=0\\x^{2} +6x-3x-18=0\\x*(x+6)-3*(x+6)=0\\(x+6)*(x-3)=0\\\left \{ {{x+6=0} \atop {x-3=0}} \right. \\\left \{ {{x=-6} \atop {x=3}} \right. \\x_1=-6, x_2=3$4) $x^{2} -5x-24=0\\x^{2} +3x-8x-24=0\\x*(x+3)-8*(x+3)=0\\(x+3)*(x-8)=0\\\left \{ {{x+3=0} \atop {x-8=0}} \right. \\\left \{ {{x=-3} \atop {x=8}} \right. \\x_1=-3, x_2=8$

5) $x^{2} +9x-22=0\\x^{2} +11x-2x-22=0\\x*(x+11)-2*(x+11)=0\\(x+11)*(x-2)=0\\\left \{ {{x+11=0} \atop {x-2=0}} \right. \\\left \{ {{x=-11} \atop {x=2}} \right. \\x_1=-11, x_2=2$6) $x^{2} +6x+8=0\\x^{2} +4x+2x+8=0\\x*(x+4)+2*(x+4)=0\\(x+4)*(x+2)=0\\\left \{ {{x+4=0} \atop {x+2=0}} \right. \\\left \{ {{x=-4} \atop {x=-2}} \right. \\x_1=-4, x_2=-2$