Question

given the original equation,determine the value(s) of a variable i n which there are... one solution, 2 com conj solutions, a neg solution.

i dont have examples, but can someone tell me what i should do?

Answer 1

for quadratics

Answer 2

For quadratics, how can we get only one solution??

We get two solutions, that can be same or equal, is that what you mean?

Answer 3

i think it's asking for discriminant

Answer 4

and there's also a question like : given the discrimants, find the original equation?

Answer 5

my teacher never taught this..

Answer 6

@satellite73

Answer 7

here's an example : value{s) of k for which there is only one solution

16x^2+kx+9=0

Answer 8

See, in an equation of the form : \(ax^2 + bx + c =0\)

Discriminant (D) is given by:
\[D = b^2 - 4ac\]

Answer 9

i know how to find this one, but how would i find values that make it have complex conjugates, a neg solution, or whatever else

Answer 10

Can you find its Discriminant?

Answer 11

yes i did that^ so the answer for ONE solution would be +- 24

Answer 12

the discrimant is-552 or -600

Answer 13

+-24 to make it a perfect square

Answer 14

i mean 0

Answer 15

for one solution

Answer 16

how would i find values to make it a negative solution... for example

Answer 17

See :
\(D = b^2 - ac = k^2 - 4 \cdot 16 \cdot 9 = 0 = k^2 = 576\)

Answer 18

yup

Answer 19

So, if \(k= \pm 24\), then you will get one solution..

Answer 20

\[D = 0 \implies One \ solution \ or \ Equal \ Solution \\ D > 0 \implies Two \ Distinct \ \mathbb{R} \ Solutions \\ D < 0 \implies Two \ Complex \ Solutions\]

Answer 21

For two complex conjugate solutions, D should be less than 0..

Answer 22

oh

Answer 23

like an inequality

Answer 24

\[\implies k^2 - 576 < 0 \\ k^2 < 576 \\ k < \pm 24\]

Answer 25

oh, that's more simpler than i thought! lol

Answer 26

Choose any value of \(k\) less than 24 but greater than -24..

Answer 27

but how would i find for 2 unequal real solutions?

Answer 28

\(-24 < k < 24\) This will give you two complex..

Answer 29

For that D > 0..

Answer 30

wouldn't negative solution and complex be the same?

Answer 31

So, \(k^2 > 576\), this will give you two unequal solutions..

Answer 32

how come^?

Answer 33

yes, negative and two complex are same, if D is negative, then two complex solutions you will get..

Answer 34

\[D > 0 \\ k^2 - 576 > 0 \\ k^2 > 576\]

Answer 35

By this you will get two unequal real solutions for D > 0..

Answer 36

wouldn't that simplify to k>+-24?

Answer 37

So, for \(k >24\) and \(k< -24\) you will get two unequal real solutions..

Answer 38

2 unequal real solutions is the same as 2real irrational right?

Answer 39

real irrational??

Real are the bigger set of numbers, it includes both rational and irrational numbers..

Answer 40

wouldnt k>+-24 bring out to ---> k>-24 and K>24?

Answer 41

why would it be k<-24 and k>24

Answer 42

2 unequal real solutions or two distinct solutions we say..

But yes, if solution is like : \(2+ \sqrt{3}\) and \(2 - \sqrt{3}\), these are real irrational here..

Answer 43

Choose a number lesser than -24..

Answer 44

I choose : \(k=-25\)..

Find D for it, is it coming positive?

Answer 45

uhyes

Answer 46

ah okay, i guess that's how it is..

Answer 47

thanks, my whole class didn't understand the test question..

Answer 48

So, \(k<-24\) will give you positive D.. Similarly, \(k>24\) will give you positive D.

Squaring of \(k\) is making that to happen that way, getting?

Answer 49

yes

Answer 50

it makes sense :D

Answer 51

If you have doubt anywhere, then just take a value for \(k\) then check for D.
If it is 0, then one solution.
If D < 0, then two complex
if D > 0, then two unequal real..

Answer 52

All good?

Answer 53

i also have like 2 more questions D:

Answer 54

that i want to clear up

Answer 55

Sure, go ahead.. :)

Answer 56

so if you were to be given a discriminant, is it possible to find the original equations?

Answer 57

I will arrange help for you as I don't have much time to stay here more, don't worry you will get help here always..

Answer 58

cause it's possible if you were given the roots

Answer 59

With only given D, it is difficult to say..

Answer 60

ah okay. thanks !!!

Answer 61

i think that's all :o

Answer 62

Do you have any example with you?

Answer 63

@apoorvk or @iGreen can we have your eyes here? Not physically though..!! :P

Answer 64

well i wasn't sure, because my friend said that there were questions like : he was given a discrimant and had to find the original equation... but i wasn't sure if he was given extra info...

Answer 65

but he probably over exaggerated

Answer 66

You need it now or you can wait?
If nothing comes here to help you then I can but tomorrow at the same time or sometime earlier than this, but tomorrow..

Answer 67

Is that okay>

Answer 68

nah i have until tomorrow ::)

Answer 69

but thanks so much. if you could figure it out. tell me!

Answer 70

Sure.. :)

Answer 71

Check this links to practice questions:
http://2012books.lardbucket.org/books/beginning-algebra/s12-04-guidelines-for-solving-quadrat.html

Answer 72

See, I have broken my head on your question, but according to me, if only D is given, you cannot figure out its Original Equation..

Answer 73

Because for you can get same D for two or more equations..