I really need help with a quadratic question.
So the problem is
kx^2+[k+2]x-3=0 has roots which are real and positive. Find the possible values that k may have.
The answer is -8+ √60<k<0
I get why they got -8+ √60 but I don't understand the 0. I thought it would be
-8+ √60 <k < -8- √60
I will very much appreciate it if anyone could help me with this problem!
using the quadratic formula, we have
x = [-(k+2)±√((k+2)^2 + 12k)]/2k
For the roots to be real, we need
(k+2)^2 + 12k >= 0
-8-√60 <= k <= -8+√60
Now, we want the roots to be positive, so we need
[-(k+2)±√((k+2)^2 + 12k)]/2k > 0
Tough question!
roots are:
x = ( -k-2 ± √(k+2)^2 - 4(k)(-3) )/2k
= ( -k-2 ±√(k^2 + 4k + 4 + 12k))/2k
it said: both x's are positive, so
( -k-2 + √(k^2 + 4k + 4 + 12k))/2k > 0
assuming we started with a positive x^2 term, so k > 0
-k-2 + √(k^2 + 4k + 4 + 12k) > 0
√(k^2 + 16k + 4) > k+2
square both sides:
k^2 + 16k + 4 > k^2 + 4k + 4
12k > -4
k > -1/3
or
-k-2 - √(k^2 + 4k + 4 + 12k) > 0
-k-2 > √(k^2 + 4k + 4 + 12k)
square both sides
k^2 + 4k + 4 > k^2 + 16k + 4
0 > 12k
k < 0
for real roots:
b^2 - 4ac > 0
(k+2)^2 -4k(-3) > 0
k^2 + 4k + 4 + 12k > 0
k^2 + 16k + 64> -4 + 64 <----- completed the square
(k+8)^2 > 60
±(k+8) > √60
k+8 > √60 , k > -8+√60 , ( k > appr -.25)
or
-k-8 > √60
-k > 8 + √60
k < -8 - √60 , (k < appr -15.75)
so to be real, -15.75
so testing (since we squared, all answers must be tested), we have 4 conditions
k < 0
k > -1/3
k > -8 + √60 or k > appr -.25
k < -8 - √60 or k < -15.75
and using Wolfram , we can see that
indeed -8+√60 < k < 0
http://www.wolframalpha.com/input/?i=solve+kx%5E2%2B%5Bk%2B2%5Dx-3%3D0+for+k%3D+-.2
change the value of k to something like k = 2 and you will see one positive and one negative root
try different values of k
To find the possible values for k in the given quadratic equation, you need to consider two conditions:
1. The discriminant, b^2 - 4ac, must be greater than or equal to zero for the equation to have real roots.
2. Both roots must be positive.
Let's consider these conditions one by one.
1. Discriminant condition:
The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac. In this case, kx^2 + (k+2)x - 3 = 0, so a = k, b = (k+2), and c = -3. The discriminant is:
(k+2)^2 - 4(k)(-3) = k^2 + 4k + 4 + 12k = k^2 + 16k + 4
For real roots, the discriminant should be greater than or equal to zero:
k^2 + 16k + 4 ≥ 0
2. Positive roots condition:
To find the positive roots, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, the ± symbol implies that we consider both the positive and negative roots. However, since we're looking for positive roots, we only consider the positive value:
x = (-b + sqrt(b^2 - 4ac)) / 2a
Substituting the values from our equation, we have:
x = (-(k+2) + sqrt((k+2)^2 - 4(k)(-3))) / 2k
Simplifying this expression, we find:
x = (-k - 2 + sqrt(k^2 + 4k + 4 + 12k)) / 2k
= (-k - 2 + sqrt(k^2 + 16k + 4)) / 2k
= (-k - 2 + sqrt((k^2 + 16k + 4))/ 2k
Now, let's analyze the solution.
For positive roots, the numerator (-k - 2 + sqrt((k^2 + 16k + 4)) must be greater than zero:
-k - 2 + sqrt((k^2 + 16k + 4)) > 0
Re-arranging the inequality, we have:
sqrt((k^2 + 16k + 4)) > k + 2
Squaring both sides of the inequality, we get:
k^2 + 16k + 4 > (k + 2)^2
k^2 + 16k + 4 > k^2 + 4k + 4
16k > 4k
This inequality simplifies to:
12k > 0,
k > 0
Thus, we know that k must be greater than 0 for both roots to be positive.
Now, let's consider the discriminant condition:
k^2 + 16k + 4 ≥ 0
This quadratic inequality can be solved by factoring or using the quadratic formula. However, since the discriminant is positive, we know that the roots of this equation are real.
By solving the quadratic inequality, we find:
(k - (-8 - sqrt(60)))(k - (-8 + sqrt(60))) ≥ 0
(k + 8 + sqrt(60))(k + 8 - sqrt(60)) ≥ 0
This inequality tells us that either both factors are positive or both factors are negative.
Case 1: Both factors are positive
k + 8 + sqrt(60) > 0 --(1)
k + 8 - sqrt(60) > 0 --(2)
Solving inequalities (1) and (2), we find:
k > -8 - sqrt(60)
k > -8 + sqrt(60)
Case 2: Both factors are negative
k + 8 + sqrt(60) < 0 --(3)
k + 8 - sqrt(60) < 0 --(4)
Solving inequalities (3) and (4), we get:
k < -8 - sqrt(60)
k < -8 + sqrt(60)
Therefore, combining both cases, we have:
-8 - sqrt(60) < k < -8 + sqrt(60)
which can be further simplified to:
-8 + sqrt(60) < k < -8 - sqrt(60) (after rearranging terms)
So, the possible values for k are:
-8 + sqrt(60) < k < -8 - sqrt(60)
It seems like you made an error in your calculation, assuming it would be -8 + sqrt(60) < k < -8 - sqrt(60). However, the correct range is -8 + sqrt(60) < k < -8 - sqrt(60).