Mathematics, 23.09.2019 14:00 cjdaly123

Solve the equation using the quadratic formula. x squared minus 4 x plus 3 equals 0 (1 point)x equals 2 semicolon x equals 4x equals 0 semicolon x equals 4x equals 1 semicolon x equals 3x equals negative 1 semicolon x equals 92. solve the equation using the quadratic formula. x squared plus 10 x equals negative 25 (1 point)x equals 5x equals negative 5x equals 15x equals 253. using the discriminant, determine the number of real solutions. negative 4 x squared plus 20 x minus 25 equals 0 (1 point)no real solutionsone real solutiontwo real solutions4. using the discriminant, determine the number of real solutions.2 x squared plus 7 x minus 15 equals 0 (1 point)no real solutionsone real solutiontwo real solutions5. using the discriminant, determine the number of real solutions. negative 2 x squared plus x minus 28 equals 0 (1 point)no real solutionsone real solutiontwo real solutions6. if the discriminant of a quadratic is zero, determine the number of real solutions. (1 point)no real solutionsone real solutiontwo real solutions7. solve the equation using the quadratic formula.3 x squared equals 2 left-parenthesis 2 x plus 1 right-parenthesis (1 point)x equals startfraction 3 plus minus startroot of 13 endroot over 6 endfractionx equals startfraction 2 plus minus startroot of 10 endroot over 3 endfraction x equals one-sixth semicolon x equals one-halfx equals 3 semicolon x equals negative 18. solve the equation using any technique. x squared plus 8 x plus 12 equals 0 (1 point)x equals negative 4 semicolon x equals 4x equals negative 6 semicolon x equals negative 2x equals 3 semicolon x equals 7x equals 6 semicolon x equals 89. solve the equation using any technique.6 x squared minus 5 x minus 1 equals 0 (1 point)x equals one-fifth semicolon x equals one thirdx equals 2 semicolon x equals 6x equals 3 semicolon x equals negative one-halfx equals negative one-sixth semicolon x equals 110. solve the equation using any technique. x squared equals 3 x minus 1 (1 point)x equals startfraction 2 plus minus startroot 7 endroot over 3 endfractionx equals startfraction 1 plus minus startroot 3 endroot over 2 endfractionx equals startfraction 3 plus minus startroot 5 endroot over 2 endfractionx equals startfraction negative 2 plus minus 2 startroot 3 endroot over 3 endfraction

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Answers

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. $x=\frac{2\pm \sqrt{10}}{3}$
8. x= -6, x = -2
9. x = -1/6, x=1
10. $x=\frac{3\pm \sqrt{5}}{2}$

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

Answer from: Quest

more detail

step-by-step explanation:

Answer from: Quest

answer: 35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %, p(e) = 27 % and $p(l\cap e)$ = 4 %

and, we have to find out, $p(l\cup e)$

but, we know that,

$p(l\cup e)= p(l) + p(e) - p(l\cap e)$

which is the required equation that will use to find the probability of a late departure or an early departure.

$p(l\cup e)= 12 + 27 - 4$

$p(l\cup e)= 35 %$

therefore, the probability of a late departure or an early arrival = 35 %

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