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Yuliya22 [10]
3 years ago
5

The roots of a quadratic equation ax2+bx+c=0 are 3+sqrt2 and 3−sqrt2. Find the values of b and c assuming that a=1.

Mathematics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

b=-6\\c=7

Step-by-step explanation:

we know that

The general equation of a quadratic function in factored form is equal to

y=a(x-x_1)(x-x_2)

where

a is a coefficient of the leading term

x_1 and x_2 are the roots

we have

a=1\\x_1=3+\sqrt{2}\\x_2=3-\sqrt{2}

substitute

y=(1)(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))

Applying the distributive property convert to expanded form

y=x^2-x(3-\sqrt{2})-x(3+\sqrt{2})+(3+\sqrt{2})(3-\sqrt{2})

y=x^2-(3x-x\sqrt{2})-(3x+x\sqrt{2})+(9-2)

y=x^2-3x+x\sqrt{2}-3x-x\sqrt{2}+7

y=x^2-6x+7

therefore

b=-6\\c=7

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Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = P(Laptop) = p_{X} = \frac{4}{9}

The probability of selecting a desktop is = P(Desktop) = p_{Y} = \frac{5}{9}

Then both X and Y follows Binomial distribution.

X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})

The probability function of a binomial distribution is:

P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: {n\choose k}=\frac{n!}{k!(n-k)!}

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\  =0.304832\\\approx0.305

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

P(Y\geq 3)=1-P(Y

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0
3 years ago
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