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slega [8]
3 years ago
7

Use the discriminant to describe the roots of each equation. Then select the best description?

Mathematics
2 answers:
Nitella [24]3 years ago
5 0

Answer:

The roots are real and irrational

Step-by-step explanation:

* Lets explain what is the discriminant

- In the quadratic equation ax² + bx + c = 0, the roots of the

 equation has three cases:

1- Two different real roots

2- One real root or two equal real roots

3- No real roots means imaginary roots

- All of these cases depend on the value of a , b , c

∵ The rule of the finding the roots is

   x = [-b ± √(b² - 4ac)]/2a

- The effective term is √(b² - 4ac) to tell us what is the types of the root

# If the value under the root b² - 4ac positive means greater than 0

∴ There are two different real roots

# If the value under the root b² - 4ac = 0

∴ There are two equal real roots means one real root

# If the value under the root b² - 4ac negative means smaller than 0

∴ There is real roots but the roots will be imaginary roots

∴ We use the discriminant to describe the roots

* Lets use it to check the roots of our problem

∵ x² - 5x - 4 = 0

∴ a = 1 , b = -5 , c = -4

∵ Δ = b² - 4ac

∴ Δ = (-5)² - 4(1)(-4) = 25 + 16 = 41

∵ 41 > 0

∴ The roots of the equation are two different real roots

∵ √41 is irrational number

∴ The roots are real and irrational

* Lets check that by solving the equation

∵ x = [-(-5) ± √41]/2(1) = [5 ± √41]/2

∴ x = [5+√41]/2 , x = [5-√41]/2 ⇒ both real and irrational

Musya8 [376]3 years ago
5 0

Answer:

b

Step-by-step explanation:

Calculate the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and irrational

• If b² - 4ac > 0 and a perfect square, roots are real and rational

• If b² - 4ac = 0 then roots are equal, double root

• If b² - 4ac < 0 then roots are not real, imaginary roots

For x² - 5x - 4 = 0

with a = 1, b = - 5 and c = - 4, then

b² - 4ac

= (- 5)² - (4 × 1 × - 4)

= 25 + 16

= 41

Since b² - 4ac > 0 then roots are real and irrational

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An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

7 0
3 years ago
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