Strange quadratic

I'm stuck on this strange question:

kx^2 - (3k+2)x + (k+3) = 0 , has roots A and B.

i) Show that the roots are real and distinct for all numbers k,

ii) If A= 2/B, find K

iii) Given this value of K, without solving the equation, find the equation whose roots are -A^2 and -B^2.

I'll be honest, i've been on the first part for 2 hours now and I have no clue where to go. I've tried to use b^2 = 4ac but I can't get anywhere. Any people willing to help?

Thanks in advance!

By the way, my maths skills are only up to A Level.

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Reply 1

xAikx

Original post by BanglaBOSS

b^2 - 4ac > 0

Reply 2

xAikx

Original post by BanglaBOSS

You need to plug in the coefficients into that and then you get a different quadratic involving k.

Reply 3

BanglaBOSS

You need to plug in the coefficients into that and then you get a different quadratic involving k.

Thanks for the speedy reply. I solved b2 - 4ac for k, but what does this K mean? And what can I do with it to find out if b2 - 4ac is bigger, smaller or equal to 0?

Reply 4

BanglaBOSS

Sorry, i actually incorrectly solved for K.

Reply 5

BanglaBOSS

I used b^2 - 4ac, and i've got it down to 5k^2 + 4.

Reply 6

AndreiMiron

Original post by BanglaBOSS

I used b^2 - 4ac, and i've got it down to 5k^2 + 4.

and how's this expression ? can it be less than 0 ?

Reply 7

BanglaBOSS

That's the problem. The question says f(x) has two distinct roots, but if i solve b^2 - 4ac for K, i get a negative number.

Reply 8

BanglaBOSS

Original post by xAikx

You need to plug in the coefficients into that and then you get a different quadratic involving k.

I'm down to 5k^2 + 4. Solving for k gets me a negatice number, but the original equation is meant to have 2 distinct roots :redface:

Reply 9

AndreiMiron

Original post by BanglaBOSS

That's the problem. The question says f(x) has two distinct roots, but if i solve b^2 - 4ac for K, i get a negative number.

I don't think you do . Let's just break it down

first term 5k^2 can this term ever be less than 0 ? for any real k
second term 4 which is obviously positive regardless of k

Reply 10

BanglaBOSS

So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?

Reply 11

AndreiMiron

Original post by BanglaBOSS

So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?

for this to happen delta(b^2 - 4ac) must be strictly greater than 0 ( real = delta is greater or equal to 0 , distinct = delta is not 0)

Reply 12

BanglaBOSS

That means my delta was wrong. Would you be able to work out the delta please?

Reply 13

AndreiMiron

Original post by BanglaBOSS

That means my delta was wrong. Would you be able to work out the delta please?

your delta is right 5k^2 +4

Reply 14

AmmarTa

Original post by BanglaBOSS

So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?

Solve in terms of b^2-4ac > 0 showing that it has 2 distinct real solutions
If it comes out with (3k+2)^2 - 4k(k+3) you get (9k^2+12k+4) - (4k^2+12k) which is 5k^2 + 4
Since k^2 is guaranteed to be positive even if k is negative (look up squares mate) then the discriminant will always be > 0 as even if k = 0 then the +4 constant means the min value will be 4 :smile:

(edited 6 years ago)

Reply 15

BanglaBOSS

Original post by AmmarTa

Thanks for the really detailed reply :smile:

, where I am getting stuck is where K^2 > -4/5

You can't square root negatives right?

Reply 16

AndreiMiron

Original post by BanglaBOSS

Thanks for the really detailed reply :smile:

, where I am getting stuck is where K^2 > -4/5

You can't square root negatives right?

out of context : you can indeed root square negative numbers but the result will be a complex number
but here k is real so the square is obviously greater than any negative number

Reply 17

AmmarTa

Original post by BanglaBOSS

Thanks for the really detailed reply :smile:

, where I am getting stuck is where K^2 > -4/5

You can't square root negatives right?

As far as A Level maths goes (further covers it), no. You only really deal with "real" and "natural" numbers and square rooting a negative doesn't give you an answer that's either of those (real is just a regular number, can be any number; natural is all positive integers starting at 1, it's the numbers that would've been used for counting for maths' most original and primitive uses). I'm interested in where you got k^2 > -4/5 from?

Reply 18

BanglaBOSS

Thanks for the replies guys! I used b^2 - 4ac to get 5k^2 +4. As f(x) has 2 roots, 5k^2 + 4 > 0. I then subtracted 4 and divided by 5 on both sides, to get k^2 > -4/5.

Reply 19

AmmarTa

As far as I can see, to solve part 2, you'd work out the 2 roots using the quadratic formula [(-b +- root(b^2 - 4ac))/2a] using your coefficients of a, b and c:
a = k, b = 3k + 2, c = k + 3
x = [-(3k+2) + or - root(5k^2+4)]/2k
We'll say that the positive form is A and the negative form is B:
A = (-(3k+2) + root(5k^2+4)) / 2k
B = (-(3k+2) - root(5k^2+4)) / 2k
Therefore since A = 2 / B we can set this to be true
(-(3k+2) + root(5k^2+4)) / 2k = 2 / ((-(3k+2) - root(5k^2+4)) / 2k)
From here, there's a few ways to solve it - we could painfully multiply the LHS by the denominator on the RHS to get an equation involving the k terms on once side and simply 2 on the other side. I'm not typing that monstrosity of an equation so instead I'll just tell you k = 3 (I'm sure you can multiply and expand brackets).