Exercises 18-20: In each of these problems, an equation and one of its roots are given. Find:
a the value of k
b the second root
18.5 is a root of x^2- 7x + k = 0
19. 5 is the root of x^2-3x+k=0
20.7 is the root of x^2-3x= -k
Please help! It’s three questions

(18) a. The value of k is 10.

b. The second root is 2.

(19) a. The value of k is -10.

b. The second root is -2.

(20) a. The value of k is -28.

b. The second root is -4.

Step-by-step explanation

18. One of the roots is 5, :. x = 5

Substitute x = 5 into the quadratic equation to obtain the value of k.

x² - 7x + k = 0

5² - 7(5) + k = 0

25 - 35 + k = 0

-10 + k = 0

k = 0 + 10 = 10

Factorise the quadratic equation to obtain the value of the second root.

x² - 7x + 10 = 0

x² - 5x - 2x + 10 = 0

(x² - 5x) - (2x + 10) = 0

x(x - 5) -2(x - 5) = 0

(x - 5)(x - 2) = 0

x = 5 or 2

The second root is 2.

19. One of the roots is 5, :. x = 5

Substitute x = 5 into the quadratic equation to obtain the value of k.

x² - 3x + k = 0

5² - 3(5) + k = 0

25 - 15 + k = 0

10 + k = 0

k = 0 - 10 = -10

Factorise the quadratic equation to obtain the value of the second root.

x² - 3x + (-10) = 0

x² - 3x - 10 = 0

x² - 5x + 2x - 10 = 0

(x² - 5x) + (2x - 10) = 0

x(x - 5) +2(x - 5) = 0

(x - 5)(x + 2) = 0

x = 5 or -2

The second root is -2.

20. One of the roots is 7, :. x = 7

Substitute the value of x into the quadratic equation to obtain the value of k.

x² - 3x = -k

7² - 3(7) = -k

49 - 21 = -k

28 = -k

k = -28

Factorise the quadratic equation to obtain the value of the second root.

x² - 3x = -k

x² - 3x = 28

x² - 3x - 28 = 0

x² - 7x + 4x - 28 = 0

(x² - 7x) + (4x - 28) = 0

x(x - 7) +4(x - 7) = 0

(x - 7)(x + 4) = 0

x = 7 or -4

The second root is -4.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

answer:

a=9

step-by-step explanation: lmk if you need a explanation