Question

Find the number of real solutions depending on values of k in 8^[x(x+k)] = [1/16]^(27/4)

Answer 1

I got it down to the equation x^2 + kx + 9 = 0

Answer 2

8^[x(x+k)] = [1/16]^(27/4)

(2^3)^[x(x+k)] = [2^(-4)]^(27/4)

2^[3x(x+k)] = 2^(-4*27/4)

3x(x+k) = -27

3x^2 + 3k = -27

3x^2 + 3k + 27 = 0

The last equation above will have real solutions when \(\Large D \ge 0\)

D is the discriminant is given by

D = b^2 - 4ac

So 3x^2 + 3k + 27 = 0 will have real solutions when \(\Large b^2 - 4ac \ge 0\)

In this case, a = 3, b = 3 and c = 27.

Answer 3

oops, i mean

a =3, b = 3k, and c = 27

so

\(\Large b^2 - 4ac \ge 0\)

\(\Large (3k)^2 - 4(3)(27) \ge 0\)

keep going to solve for k

Answer 4

I know you get k^2 > 36

Answer 5

good, you're very close

Answer 6

\[\left| k \right| > 6\]

Answer 7

you can write the answer as

\[\Large |k| \ge 6\]

or as

\[\Large k \le -6 \ \text{or} \ k \ge 6\]

Answer 8

But the question isn't asking me to solve for k. It says "find the number of real solutions depending on values of k"

Answer 9

Does it give you values of k to use?

Answer 10

no. That's why I was slightly confused. I heard from a friend that the answer is supposed to be 2 real values. But I'm not sure why.

Answer 11

Maybe there's more context elsewhere?

Answer 12

Oh wait I think I have an idea:

Answer 13

So above the part of the sheet we got with the questions, there is some information-
it talks about the discriminant, and how the values of the discriminant affect how many roots there are and the nature of the roots
it says that if b^2 - 4ac > 0, there are two real roots.
So would the answer be that if k≤−6 or k≥6, then there are two real roots?

Answer 14

if you mean two distinct real roots, then yes b^2 - 4ac > 0

which means that if

k < -6 or k > 6

then the original equation has 2 distinct real roots

Answer 15

i think that's what she wants. can you help me with another one?

Answer 16

(and thank you)

Answer 17

sure, what is it

Answer 18

should i post it within this question?

Answer 19

either way works

Answer 20

Find all values of k for which roots of the equation: \[5^{x(x+1)}\times25^{[k(k-1)]/[2]} = \sqrt{5^{x ^{2}}}\times125^{(kx+k+1)/(2)}\] satisfy the restriction: 1/x1 + 1/x2 > 0

Answer 21

(satisfy the restriction: \[1/x _{1} + 1/x _{2} > 0\]

Answer 22

\[\Large \frac{1}{x_{1}}+\frac{1}{x_{2}} > 0\]

\[\Large \frac{x_{2}}{x_{1}x_{2}}+\frac{x_{1}}{x_{1}x_{2}} > 0\]

\[\Large \frac{x_{2}+x_{1}}{x_{1}x_{2}} > 0\]

\[\Large \frac{x_{1}+x_{2}}{x_{1}x_{2}} > 0\]

So

\[\Large \frac{1}{x_{1}}+\frac{1}{x_{2}} > 0\]

is equivalent to

\[\Large \frac{x_{1}+x_{2}}{x_{1}x_{2}} > 0\]

Answer 23

5^[x(x+1)]*25^[k(k-1)/2] = sqrt(5^(x^2))*125^[(kx+k+1)/(2)]


5^[x(x+1)]*(5^2)^[k(k-1)/2] = (5^(x^2))^(1/2)*(5^3)^[(kx+k+1)/(2)]


5^[x(x+1)]*5^[2k(k-1)/2] = 5^((1/2)*x^2)*5^[3(kx+k+1)/(2)]


5^[x(x+1)+2k(k-1)/2] = 5^((1/2)*x^2)+3(kx+k+1)/(2)]


x(x+1)+2k(k-1)/2 = (1/2)*x^2+3(kx+k+1)/(2)


x^2 + x + k^2 - k = (1/2)*x^2 + (3kx + 3k + 3)/2


2x^2 + 2x + 2k^2 - 2k = x^2 + 3kx + 3k + 3


2x^2 + 2x + 2k^2 - 2k - x^2 - 3kx - 3k - 3 = 0


x^2 + (2 - 3k)x + (2k^2 - 5k - 3) = 0


In this case, a = 1, b = 2 - 3k, and c = 2k^2 - 5k - 3


Solve for x.


x = (-b+-sqrt(b^2-4ac))/(2a)


x = (-(2 - 3k)+-sqrt((2 - 3k)^2-4*1*(2k^2 - 5k - 3)))/(2*1)


x = (-2 + 3k+-sqrt(4 - 12k + 9k^2-8k^2 + 20k + 12))/(2*1)


x = (-2 + 3k+-sqrt(k^2 + 8k + 16))/(2)


x = (-2 + 3k+-sqrt((k+4)^2))/(2)


x = (-2 + 3k+-|k+4|)/(2)


x = (-2 + 3k+|k+4|)/(2) or x = (-2 + 3k - |k+4|)/(2)


So the two solutions for the original equation are x = (-2 + 3k+|k+4|)/(2) or x = (-2 + 3k - |k+4|)/(2)


Since we have a +/-, this takes care of the absolute value, so we can say that the two solutions are


x = (-2 + 3k+(k+4))/(2) or x = (-2 + 3k - (k+4))/(2)


x = (4k+2)/(2) or x = (2k - 6)/(2)


x = 2k + 1 or x = k - 3


So we can say


x1 = 2k + 1, x = k - 3


But above we know that


(x1 + x2)/(x1*x2) > 0


So this really means


(2k+1 + k -3)/((2k+1)(k-3)) > 0


(3k-2)/((2k+1)(k-3)) > 0


Keep going to solve for k

Answer 24

so i have [3k - 2] / [(2k+1)(k-3)] > 0

Answer 25

I'm actually not sure what to do from there

Answer 26

I can't multiply both sides by 2k^2 - 5k - 3 because I could be multiplying by zero

Answer 27

look for the zeros of [3k - 2] / [(2k+1)(k-3)]

also, look for the vertical asympotes of [3k - 2] / [(2k+1)(k-3)]

Answer 28

Are you telling me to graph it...?

Answer 29

you can if you want, but you can also determine this info without graphing

Answer 30

How? (also I haven't learned asymptotes yet)

Answer 31

find the values of k which make the denominator zero

Answer 32

k = -1/2 or k = 3

Answer 33

What do I do with those values? The denominator can't be zero

Answer 34

3k - 2 = 0

k = 2/3

So we must take this value into consideration as well

Answer 35

so set up a number line with the values, k = -1/2, k = 3 and k = 2/3 on it

Answer 36

then test values between these numbers to see where [3k - 2] / [(2k+1)(k-3)] is positive

Answer 37

why only between?

Answer 38

because the actual values themselves either give 0 or give you a division by zero error

Answer 39

but can't you go past the values? for example, test k = 9 or test k = -7

Answer 40

you can test as many values as you want, but it only makes sense to test values from each region on the number line

Answer 41

okay so -1/4 works, 1/3 works, and 1 does not work.

Answer 42

good

Answer 43

then what?

Answer 44

these values represent intervals, you're missing one interval though

Answer 45

Which one? I have the interval between 0 and -1/2, 0 and 2/3, and 2/3 and 3

Answer 46

how about from 3 to infinity?

Answer 47

wait but i asked that. i asked if you should go beyond the limits of -1/2 and 3 and you said no. then do i also need to test something between -1/2 and - infinity?

Answer 48

("but can't you go past the values? for example, test k = 9 or test k = -7
15 minutes ago"

Answer 49

why not? they're test values. If they fail the test, then you ignore that interval

Answer 50

-5 doesn't work but 6 works. so the ones that work are between -1/2 and 0, 0 and 2/3, and 3 and infinity

Answer 51

good, those are the intervals where k makes [3k - 2] / [(2k+1)(k-3)] positive

Answer 52

oh so is that the answer? how would i formate something like that?

Answer 53

*format

Answer 54

you can use interval notation, or say k > 3 for the interval from 3 to infinity for instance

Answer 55

we haven't learned/worked with interval notation. could i say: