Answer: 1: x=3, x=1
2: x= -5
3: There are 2 real solutions.
4: There are 2 real solutions.
5: There are no real solutions.
6. There is 1 real solution.
7.
8. x= -6, x = -2
9. x = -1/6, x=1
10.
Explanation:
1. The quadratic formula is
Substituting our known information we have:
2. Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:
3. The discriminant is b²-4ac. For this problem, that is 20²-4(-4)(25)=400--400=800. Since this is greater than 0, there are 2 real solutions.
4. The discriminant in this problem is 7²-4(2)(-15)=49--120=49+120=169. This is greater than 0, so there are 2 real solutions.
5. The discriminant in this problem is 1²-4(-2)(-28)=1-224=-223. Since this is less than 0, there are no real solutions.
6. If the discriminant of a quadratic is 0, then by definition there is 1 real solution.
7. Rewriting the quadratic we have 3x²-4x-2=0. Using the quadratic formula we have:
8. Factoring this trinomial we want factors of 12 that sum to 8. 6*2 = 12 and 6+2=8, so those are our factors. This gives us:
(x+6)(x+2)=0
Using the zero product property we know that either x+6=0 or x+2=0. Solving these equations we get x= -6 or x= -2.
9. Factoring this trinomial we want factors of 6(-1)=-6 that sum to -5. (-6)(1)=-6 and -6+1=-5, so this is how we "split up" the x term:
6x²-6x+1x-1=0
We group together the first two and the last two terms:
(6x²-6x)+(1x-1)=0
Factor the GCF out of each group. In the first group, that is 6x:
6x(x-1)+(1x-1)=0
In the second group, the GCF is 1:
6x(x-1)+1(x-1)=0
Both terms have a factor of (x-1), so we can factor it out:
(x-1)(6x+1)=0
Using the zero product property, we know either x-1=0 or 6x+1=0. Solving these equations we get x=1 or x=-1/6.
10. Substituting our information into the quadratic formula we get:
Step-by-step explanation:
![z= \rho [\cos \theta + i \sin \theta]](https://tex.z-dn.net/?f=z%3D%20%5Crho%20%5B%5Ccos%20%5Ctheta%20%2B%20i%20%5Csin%20%5Ctheta%5D)
, can be calculated as follows:![z^n = \rho^n [\cos n\theta + i \sin n\theta ]](https://tex.z-dn.net/?f=z%5En%20%3D%20%5Crho%5En%20%5B%5Ccos%20n%5Ctheta%20%2B%20i%20%5Csin%20n%5Ctheta%20%5D)
is equal to![z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20%5Crho%5E3%20%5B%5Ccos%203%20%5Ctheta%20%2B%20i%20%5Csin%203%20%5Ctheta%20%5D%20%3D%20%285%5E3%29%20%5B%5Ccos%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%2B%20i%20%5Csin%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%5D)
(1)
, (1) becomes![z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20125%20%5B%5Ccos%20270%5E%7B%5Ccirc%7D%20%2B%20i%20%5Csin%20270%5E%7B%5Ccirc%7D%20%5D)