Solve x2 − 8x + 5 = 0 using the completing-the-square method. (2 points)

Mathematics

1 answer:

RoseWind [281]3 years ago

8 0

Answer:

$x = 4 + \sqrt{11}, \: \: x = 4 - \sqrt{11}$

Step-by-step explanation:

${x}^{2} -8x + 5 = 0 \\ \\ {x}^{2} - 8x + {4}^{2} = - 5 + {4}^{2} \\ \\ {(x - 4)}^{2} = - 5 + 16 \\ \\ {(x - 4)}^{2} = 11 \\ \\ x - 4 = \pm \sqrt{11} \\ \\ x = 4 \pm \sqrt{11} \\ \\ x = 4 + \sqrt{11}, \: \: x = 4 - \sqrt{11}$

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EXERCISE

Olin [163]

Answer:

1)The probability that the ball drawn is not red is 6 /14

2)Probability of getting either a prime or a multiple of 3 is 7/10

3)Probability of drawing three balls without replacement such that all three are Same colour is 0.0747

Probability of drawing three balls without replacement such that all three are different colour is 0.5

Step-by-step explanation:

1) A ball is drawn randomly from a bag containing 8 red, 4 blue and 2 green identical balls. Find the probability that the ball drawn is not red.

No. of red balls = 8

No. of blue balls = 4

No. of green balls = 2

Total no. of balls = 8+4+2=14

P(Getting red bal)= $\frac{8}{14}$

P( ball drawn is not red) = $1 - \frac{8}{14}=\frac{6}{14}$

So,the probability that the ball drawn is not red is 6 /14

2)What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?

Numbers : 1 ,2,3,4,5,6,7,8,9,10

Prime numbers: 2,3,5,7 = 4

Multiple of 3 = 3,6,9 =3

So, Probability of getting prime number = $\frac{4}{10}$

Probability of getting multiple of 3 = $\frac{3}{10}$

So, Probability of getting either a prime or a multiple of 3= $\frac{4}{10}+\frac{3}{10}=\frac{7}{10}$

Hence Probability of getting either a prime or a multiple of 3 is 7/10

3)A bag contains four red balls, five blue balls and six green balls. Three balls are drawn from the bag at random without replacement. What is the probability that all three are:

The same colour

Different colour

No. of red balls = 4

No. of blue balls = 5

No. of green balls = 6

Total balls = 15

Case 1 :all three are Same colour

So, probability of drawing three balls without replacement such that all three are Same colour = $\frac{4}{15} \times \frac{3}{14} \times \frac{2}{13}+\frac{5}{15} \times \frac{4}{14} \times \frac{3}{13}+\frac{6}{15} \times \frac{5}{14} \times \frac{4}{13}=0.0747$

So, probability of drawing three balls without replacement such that all three are different colour = $\frac{4}{10} \times \frac{5}{9} \times \frac{6}{8}+\frac{5}{10} \times \frac{4}{9} \times \frac{6}{8}+\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=0.5$