Question about the discriminant of a quadratic?

Question

Hi, again I am stuck on something simple and its one of them annoying prrove questions that you cant check your answer in the back of the book.

It says:

Given that k is a real constant such that 0
a) always real
b) always negative

well firstly I dont get how if the roots are real (and positive) how they can possibly be negative, but for part a at least I have done the following

Its a question about the discriminan b^2 -4ac and since b=2 a=k and c = 1-k I have done the following showing that the answer has to be greater than 0

so 2^2 - 4k(1-K) >0
so 4 - 4k + 4k^2 >0
divide the whole lot by 2
so 2 - 2k +2k^2 >0
so 2k^2 -2k > -2
so 2k(k-1) > -2
so k(k-1) > -1

but im stuck there and have no idea where to go next or how to show they are always negative

Answer 1

Hi John,



You reached a point where you're very close to solution. I think you had a small confusion with the sign of the roots: in fact they exist because the discriminant is positive (and k is positive, also), but the roots are negative.



It is as follows:



the equation is: kx^2 +2x + (1-k)



As you correctly observed, then the discriminant condition is:



2^2 - 4k(1-k) > 0



(to ensure we have two roots)



and also as you observed, this leads to:



k(k-1) > -1



until this point you're right: to ensure that TWO REAL ROOTS EXIST, the discriminant must be positive, and so k(k-1) > -1 must be verified.



Now, you know FROM THE LETTER OF THE PROBLEM that 0 < k < 1, and so -1 < k -1 < 0, and so we have:



1) k is positive and k < 1

2) (k-1) is negative and -1 < (k-1)



combining 1 and 2, you obtain: k*(k-1) is negative (positive * negative makes negative), and has absolute value < 1 (because both k and (k-1) have absolute value < 1).



Being negative and having absolute value < 1 implies that you have to be between -1 and 0, so we find that:



-1 < k*(k-1) < 0



(this condition, DEDUCED FROM THE LETTER OF THE PROBLEM)



And you had found that if k(k-1) > -1, then the discriminant is > 0



So, you proved that, if 0 < k < 1 (as the letter of the problem says), the discriminant is positive.



So you solved point "a": if 0 < k < 1, then the given equation has TWO REAL ROOTS (the discriminant is positive).





Let's go now for part "b". We have to prove that these two roots are negative. But this is very easy, because if you call D the discriminant, then these are the two roots:



R1 = (-2 - root(D)) / 2*k

R2 = (-2 + root(D)) / 2*k



R1 is clearly negative (as -2 and -root(D) are negative)



To prove that R2 is also negative, we have to prove that:



-2 + root(D) < 0



e.g. we have to prove that root(D) < 2



Remember that D = 2^2 - 4k(1-k)

and so: D = 4 - 4k(1-k)

so: D = 4(1 - k(1-k))

so: D = 4(1 + k(k-1))



and we proved that k(k-1) < 0 (negative), so we have:



1 + k(k-1) < 1 (as k(k-1) is negative)



and so: 4(1 + k(k-1)) < 4

e.g.

D < 4



as D < 4, we have root(D) < 2



and so: -2 + root(D) < 0, as we wanted to prove



we have: R2 = (-2 + root(D)) / 2k < 0



and then we proved that both R1 and R2 are negative.



At the end, this is our conclusion:



A) given 0 < k < 1, the discriminant D is strictly positive, and so TWO REAL ROOTS EXIST



B) these two real roots, R1 and R2, are both NEGATIVE



Hope it helps!

Answer 2

First of all, if the roots are "real" it doesn't follow that they are positive. Negative numbers are real, too! Non-reals are imaginary numbers, in the case of the quadratic formula when the discriminant itself is negative (which means you're finding the square root of a negative number: can't happen).



The discriminant simply tells you whether you have two real roots (the discriminant is positive) one double root (the discriminant is 0) or two imaginary roots (the discriminant is negative); as well as whether or not the roots are rational (generally, when the discriminant is a square. This gets complicated, though, when the coefficients in your original equation are not rational.)



So, to show that the roots are real, you must show that b^2-4ac is not a negative value.



For part a of your question, I would leave the constant term on the right hand side of the inequality:

2 - 2k + 2k^2 >0, since you're trying to show that the expression is greater than 0, after all. Divid through by 2:

k^2 - k + 1 >0

k^2 + 1 > k This is obviously true for all values between 0 and 1, so you have shown that the discriminant is greater than 0, and hence the roots are real.



As far as part b is concerned, you need to show that -b plus/minus root(b^2-4ac)/2a is always negative:



-2 plus/minus root(4-4*k*(1-k))/2k

-2 plus/minus root(4-4k-4k^2)/2k

-2/2k plus/minus root(4-4k-4k^2)/root(4k^2)

-2/2k plus/minus root((4-4k-4k^2)/4k^2)

-2/2k plus/minus root((1-k-k^2)/k^2)

-1/k plus/minus root(1-k-k^2)/k

(-1 plus/minus root(1-k-k^2)/k



Since k is positive, the roots will be negative when the numerator is negative. -1 minus the root of... will certainly be negative, so the question is whether -1 + root(1-k-k^2) can ever be non-negative; or, more succinctly, whether root(1-k-k^2) can ever be at least 1. It should be obvious that between 0 and 1 that k is greater than k^2, so 1 - k - k^2 will always be less than 1, and the numerator will still be negative. So, until k reaches 1, both roots of such an equation will be negative.



Sorry I couldn't make this any clearer- it's hard to type here, and there are some interseting concepts involved. Good luck!

Answer 3

Go back to your expression 4-4k+4k^2>0

Divide by 4 getting 1-k +k^2 > 0

Now since 00

This means that the roots are always real for the range of k given.



The roots of the equation are [- 2 +/- sqrt(4-4k+4k^2)]2k

= [-2 +/- 2*sqrt(1-k+k^2)]/2k = [-1 +/- sqrt(1-k+k^2)]/k

the root [-1-sqrt(1-k+k^2)]/k is clearly negative.

the root [-1+sqrt(1-k+k^2)]/k is also negative because

sqrt(1-k+k^2) is always less than 1 when 0
It has a minimum value of .75 when k=1/2 and approaches 1 (but never reaches it) when k --> 0 and k --> 1.

Answer 4

Discriminant

D = 4k^2 - 4k + 4

= [(2k)^2 - 2. 2k . 1 + 1^2 ] + 3

= (2k - 1)^2 + 3

= always positive if k is real



So the roots are always real.



Discriminant value increases if k increases

Say k = 1 - h where h is very very small

D = (2k - 1)^2 + 3

= [2(1-h) -1]^2 + 3

= [1-2h]^2 + 3

= 1 - 4h + 4h^2 + 3 ( h^2 is very very very small)

= 4 - 4h + h^2 + 3h^2 ( ignore 3h^2)

= (2 - h)^2



sqrt(D) = 2 - h



Now roots of given equation

x1 = [-2 + sqrt(D)] / 2k

x2 = [-2 - sqrt(D)] / 2k = -[2 + sqrt(D)] / 2k

By observation

x2 = negative



x1 = appx [-2 + 2 - h]/2k

= [-h/2k]

= negative



So the roots of the equation kx^2 +2x + (1-k) are:



a) always real

b) always negative

Answer 5

Bartacuba's solution was excellent. But I found an even

simpler way to do this.

Suppose kx² + 2x + 1-k = 0 (*),

where 0 < k <1.

The discriminant of this equation, d, is 4 -4k(1-k).

Since 0 < k < 1, we have 0 < k(1-k)
So d > 4 -4k >0, because 4k < 4.

Thus (*) has real roots.

Now divide (*) by k:

We have

x² + 2/k + (1-k)/k = 0. (**)

Let r1 and r2 be the roots of (**) (and therefore of (*) ).

Then (**) must factor as (x-r1)(x-r2).

Equating coefficients of like powers of x gives

r1 + r2 = -2/k, which is negative, since k is positive.

r1r2 = (1-k)/k, which is positive since 0 < k < 1.

Since r1r2 is positive, r1 and r2 must have

the same sign.

Since their sum is negative, both must be negative.



Hope that helps a bit!!

Answer 6

a) always real

You are given 0 < k < 1 and you need to prove 2^2 - 4k(1-K) >0.

Proof:

f(k) = 2^2 - 4k(1-K) = 4k^2-4k+4

f(k) is a parabola and opens up. Therefore its minimum is at k = -b/(2a) = 4/8 = 1/2

f(1/2) = 1-2+4 = 3

Since the minimum of 2^2 - 4k(1-K) is 3, 2^2 - 4k(1-K) >0 must be true.



b) always negative

f(x) = kx^2 +2x + (1-k)

Since 0 < k < 1, we can use the quadratic formula to prove b).

Proof.

root = (-2±√[4-4k(1-k)])/(2k)



Therefore, we only need to prove -2+√[4-4k(1-k)]) < 0, or-1+√[1-1k(1-k)])< 0

which is equivalent to

1-k(1-k) < 1

or -k(1-k) < 0



Since k is positive, 1-k is also positive because k < 1, -k(1-k) < 0 must be true.



END

---------

raj,

"k(k-1)>-1

so k must be between 0 and 1" is not right. Counterexample. k = 2.

Answer 7

Given that k is a real constant such that 0


a) always real

b) always negative



k(k-1)>-1

so k must be between 0 and 1

it is given k is between 0 and 1

so the roots are real

the roots are [-2+/-rt(4-4k+4k^2)]/2k

since 0
so -2+/-rt(4-4k+4k^2) is negative

and so the roots negative

Answer 8

with the roots they mean the solutions, they should be negative.



you are on the right trach, just show that the solutions (roots) are always negative...



hmmm ? the solutions are :

k1,2 = {-2 +or- sqrt(4-4k(k-1)) }/ 2k



as far as i can see you have one positive root and one negative root.