Respuesta :
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. [tex]x=\frac{2\pm \sqrt{10}}{3}[/tex]
8. x= -6, x = -2
9. x = -1/6, x=1
10. [tex]x=\frac{3\pm \sqrt{5}}{2}[/tex]
Explanation:
1. The quadratic formula is
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substituting our known information we have:
[tex]x=\frac{--4\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)} \\ \\=\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm \sqrt{4}}{2}=\frac{4\pm2}{2} \\ \\=\frac{4+2}{2},\frac{4-2}{2}=\frac{6}{2},\frac{2}{2}=3,1[/tex]
2. Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:
[tex]x=\frac{-10\pm \sqrt{10^2-4(1)(25)}}{2(1)}=\frac{-10\pm \sqrt{100-100}}{2} \\ \\=\frac{-10\pm \sqrt{0}}{2}=\frac{-10\pm0}{2}=\frac{-10}{2}=-5[/tex]
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:
[tex]x=\frac{--4\pm \sqrt{(-4)^2-4(3)(-2)}}{2(3)}=\frac{4\pm \sqrt{16--24}}{6} \\ \\=\frac{4\pm \sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}=\frac{2\pm \sqrt{10}}{3}[/tex]
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:
[tex]x=\frac{--3\pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}=\frac{3\pm \sqrt{9-4}}{2} \\ \\=\frac{3\pm \sqrt{5}}{2}[/tex]
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. [tex]x=\frac{2\pm \sqrt{10}}{3}[/tex]
8. x= -6, x = -2
9. x = -1/6, x=1
10. [tex]x=\frac{3\pm \sqrt{5}}{2}[/tex]
Explanation:
1. The quadratic formula is
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substituting our known information we have:
[tex]x=\frac{--4\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)} \\ \\=\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm \sqrt{4}}{2}=\frac{4\pm2}{2} \\ \\=\frac{4+2}{2},\frac{4-2}{2}=\frac{6}{2},\frac{2}{2}=3,1[/tex]
2. Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:
[tex]x=\frac{-10\pm \sqrt{10^2-4(1)(25)}}{2(1)}=\frac{-10\pm \sqrt{100-100}}{2} \\ \\=\frac{-10\pm \sqrt{0}}{2}=\frac{-10\pm0}{2}=\frac{-10}{2}=-5[/tex]
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:
[tex]x=\frac{--4\pm \sqrt{(-4)^2-4(3)(-2)}}{2(3)}=\frac{4\pm \sqrt{16--24}}{6} \\ \\=\frac{4\pm \sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}=\frac{2\pm \sqrt{10}}{3}[/tex]
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:
[tex]x=\frac{--3\pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}=\frac{3\pm \sqrt{9-4}}{2} \\ \\=\frac{3\pm \sqrt{5}}{2}[/tex]
