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masha68 [24]
3 years ago
8

For which value of k will the roots of the equation 2x^2-5x+k=0 be real and rational numbers

Mathematics
1 answer:
ella [17]3 years ago
5 0
The trick here is to find the discriminant, b^2-4ac, of the equation 2x^2 - 5x + k.
Here a=2, b=-5, and c=k.

So long as the discriminant is 0 or greater than 0, your roots will be real and rational numbers.

Thus, compute b^2-4ac:  (-5)^2 - 4(2)(k)

Let this equal zero and solve for k:  25=8k; k=25/8.

So long as k is 25/8 or smaller, b^2-4ac will be 0 or greater, and the roots will be real and rational.
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Answer:

Step-by-step explanation:

The standard deviation is the square root of the variance, and the variance is found by using the mean. So we will do that first. I will use the population variance as opposed to the sample variance since our set of numbers is small.

The mean: 8 + 12 + 15 + 17 + 18 = 70 and divide that by 5 to get

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s^2=\frac{(8-14)^2+(12-14)^2+(15-14)^2+(17-14)^2+(18-14)^2}{5}

Squaring this ensures us that we don't end up with zero, which would be useless.

s^2=\frac{36+4+1+9+16}{5} so

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s = 3.633

(If you do it with n-1 = 4 in the denominator of the variance, you get a standard deviation of 4.062)

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Nana76 [90]

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Step-by-step explanation:

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Step-by-step explanation:

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