Answer:
As follows,
Step-by-step explanation:
You can solve these either by factoring or using the quadratic formula or other ways,
1. x^2+2x+1=0
Factoring,
(x+1)^2=0
x+1=0
x=-1
2.y^2-y-2=0
Using quadratic formula,
(-b±√(b^2-4ac))/2a
(--1±√(-1^2-4*1*-2))/2*1
(1±√(1+8))/2
(1±√(9))/2
(1±3)/2
y=(1±3)/2
y=(1+3)/2 y=(1-3)/2
y=4/2 y=-2/2
y=2 y=-1
3.y^2+y+1=0
Using quadratic formula,
(-b±√(b^2-4ac))/2a
(-1±√(1^2-4*1*1))/2*1
(-1±√(1-4))/2
(-1±√(3))/2
No real roots
4. x^2+9x+18=0
Using quadratic formula,
(-9±√(9^2-4*1*18))/2*1
(-9±√(81-72))/2
(-9±√(9))/2
(-9±3)/2
x=(-9+3)/2 x= (-9-3)/2
x=-6/2 x=-12/2
x=-3 x=-6
5.x^2+5x+6=0
Using quadratic formula,
(-5±√(b^2-4ac))/2a
(-5±√(5^2-4*1*6))/2*1
(-5±√(25-24)/2
(-5±1)/2
x=(-5+1)/2 x=(-5-1)/2
x=-4/2 x=-6/2
x=-2 x=-3
6. 3n^2-13n-10=0
Using quadratic formula,
(-b±√(b^2-4ac))/2a
(-(-13)±√(-13^2-4*3*(-10)))/2*3
(13±√(169-(-120))/6
(13±√289)/6
The discriminant b^2−4ac>0
So there are 2 real roots.
x=(13±17)/6
x=30/6 or x=-4/6
x=5 or x=-2/3
