Предмет: Математика, автор: АринаТемникова

решите через дискриминант:
x^2+x-2=0
x^2-13x+40=0
x^2-17x+42=0
x^2-5x+6=0
x^2-7x+10=0
x^2-x-2=0

Ответы

Автор ответа: arinaari11
0
1.D=1-4*1*(-2) = 1+8=9
x1= (1+√9)/2=(1+3)/2=4/2=2
x2=(1=√9)/2=(1-3)/2=-2/2=-1
2.
Откроем модуль как число положительное, а потом как число отрицательное.
1) x ≥ 0
x² - 13x + 40 = 0
По обратной теореме Виета:
x1 + x2 = 13
x1•x2 = 40
x1 = 5
x2 = 8

2) x ≤ 0
x² + 13x + 40 = 0
x1 + x2 = -13
x1•x2 = 40
x1 = -5

x2 = -8
3. X²-17x+42=0
a=1,b=-17, c=42

D=b²-4ac, D=(-17)²-4.1.42=289-168=121,√D=√121=11
x1=(17+11)/2=28/2=14
x2=(17-11)/2=6/2=3
x1=14,x2=3 Ответ: х = -8; -5; 5; 8.
4.х2-5х+6=0
D=25-24=1
x1= (5+1)/2 = 3
x2 = (5-1) / 2=2
Ответ: х=2, х=3
5.х² - 7х + 10 = 0
х₁ + х₂ = 7
х₁ * х₂ = 10
х₁ = 5; х₂ = 2

Ответ: х₁ = 5; х₂ = 26.
D=1-4*1*(-2) = 1+8=9x1= (1+√9)/2=(1+3)/2=4/2=2
x2=(1=√9)/2=(1-3)/2=-2/2=-1
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