[Guest]

So the question goes there is a quadratic of the form X^2-ax+b=0

a and b are int

Someone copying down the question transposed the two digits of b as well as the + and - signs. Somehow one of his roots was still correct. What is this root?

now i am sure this is bad form but I havent solved it yet and I hate to admit, im stuck. I think i know the answer and it is right in front of me, but I cant mathematically prove it and I dont want to resort to guess and check. plus the answer I have makes the person that made up this problem an ******.

Enjoy

[Guest]

Clarification needed: The + and - signs were changed only for b or both a and b?

If b is rs, then is the new equation

X² - ax - sr = 0

or

X² + ax - sr = 0

Edited July 28, 2009 by DeeGee

[Pickett]

I took the OP to mean if b is rs, then the original equation was

x² - ax + rs = 0

...but he wrote it down as

x² + ax - sr = 0

(flipped the digits in b as well as the + and - sign)...

[Guest]

correct for example if the OP was

x² -2x+27=0

the new would be

x²+2x-72=0

[Guest]

Are the roots rational or is that for us to find out?

[Guest]

First of all...whosoever has posted this question...thanks to him....I would like to solve it in a much more detailed faishon....

let x & y be the actual roots of above equation....

So, x+y=a... .............(1)

xy=b=(10t + u)........(2)

By number system, any number can be written as sum of its tens & units digit...

So, t=ten's digit

u=one's digit

(eg...72=(10)x7 + 2...)

Let the common correct root be x...& z be another incorrect root..

So, xz=-(10u + t)........(3)

becoz..sign & digits get interchanged..

Now, add equn..(2)+(3)==>

x(z+y)=9(t-u)...(4)

Now, subtract equn..(3)-(2)==>

x(z-y)=-11(t+u)....(5)

Now, divide equn..(4)/(5)==>

(z+y)/(z-y)= -9(t-u)/11(t+u)...(6)

Apply..componendo ..dividendo rule on equn(6)...====>

z/y=(2t + 20u)/(-20t -2u)==-(t+10u)/(10t +u)...(7)

Compare equation, 2, 3 & 7...x=1

So, the common root is.......1....

[Guest]

the common root is rational. I thought at first about i's and realized that that is impossible. when I found the answer It turned out to be rational. Im saying this because as much as this is nice to know it really didn't help me much.

Also I figure out my mistake and found the answer but It was a little bit of brute force Im gonna try to get it completely mathematically now.

[Guest]

also I thought it could be 1 (or the other obvious guesses) but

then x^2-ax+10c+d=0

and x^2+ax-10d-c=0

plug 1 in

-a+10c+d=0

a-10d-c=0

a=10c+d

a=10d+c

0=9c-9d

c=d

x^2+ax-11c=0

x^2-ax+11c=0

add them

x^2=0

if you look if x can be zero, c must be zero making A zero and the world explodes

im pretty sure the logic is sound

[Guest]

First of all...whosoever has posted this question...thanks to him....I would like to solve it in a much more detailed faishon....

let x & y be the actual roots of above equation....

So, x+y=a... .............(1)

xy=b=(10t + u)........(2)

By number system, any number can be written as sum of its tens & units digit...

So, t=ten's digit

u=one's digit

(eg...72=(10)x7 + 2...)

Let the common correct root be x...& z be another incorrect root..

So, xz=-(10u + t)........(3)

becoz..sign & digits get interchanged..

Now, add equn..(2)+(3)==>

x(z+y)=9(t-u)...(4)

Now, subtract equn..(3)-(2)==>

x(z-y)=-11(t+u)....(5)

Now, divide equn..(4)/(5)==>

(z+y)/(z-y)= -9(t-u)/11(t+u)...(6)

Apply..componendo ..dividendo rule on equn(6)...====>

z/y=(2t + 20u)/(-20t -2u)==-(t+10u)/(10t +u)...(7)

Compare equation, 2, 3 & 7...x=1

So, the common root is.......1....

I doubt if 1 is the common root. In fact, the commmon root can not be positive!

For the equation x² - ax + b = 0, both roots must be negative!

By the way, the equation you derived for z/y in s tep 7, can be done in one step by just dividing your equations 2 and 3. I wonder how you solved eq 2,3 and 7 to get the answer as x = 1

Edited July 29, 2009 by DeeGee

[Guest]

First equation= x²-a+b

if the roots are n and m--> (x-n)(x-m)=0

n+m=-a (1)

Second equation= x²+a-b'

if the roots are n and z--> (x-n)(x-z)=0

n+z=a (2)

from (1) and (2) n+m=-n-z --> 2n=-(m+z) (3)

n*m=b=(10t+u)

n*z=-b'=-(10u+t)

Sum side by side-->

n(m+z)=9(t-u)

from(3)--> n*(-2n)=9(t-u) --> -2n²=9(t-u) -->2n²=9(u-t)

take sqrt both side-->

sqrt(2)n=3*sqrt(u-t)

for n to be an integer, sqrt(u-t) should be a multiply of sqrt(2)

so u-t is 8(2*sqrt(2)) or 2.

If u-t=8, then u=9 and t=1 -->b=91 and b'=19

but both are prime numbers, can't be equal to n*m

So u-t may only be=2 -->sqrt(2)n=3*sqrt(2) -->

n=3

[Guest]

In fact, the commmon root can not be positive!

I just realised the mistake... In fact, both roots must be positive!

So, here's the solution (the common root is not 1 indeed!)

x² - ax + rs = 0; let roots be x1 and x2

x² + ax - sr = 0; let roots be x1 and x3

x1 + x2 = a ... (1)

x1 + x3 = -a ... (2)

x1.x2 = rs

x2 = rs/x1

x1.x3 = -sr

x3 = -sr/x1

x2/x3 = -rs/sr

x3 = -(rs).x2/sr

Substitiute this is x1 + x3 = -a

and solve with x1 + x2 = a

we get

11(r+s).x1 = 9a (s-r) ...... (3)

From (1) and (2)

2x1 + x2 + x3 = 0

Then,

2x1 + rs/x1 - sr/x1 = 0

solving this we get

2x1² = 9(s-r) ... (4)

Subsitute (3) in (4) by replacing x1, we get

a² = 121(r+s)² / (18(s-r))

Since a is an integer,

(r+s)² must be divisible by 18(s-r) and the factor should be a square (1 or 4 or 9...) as we are solving for a²

This is possible only when s-r = 2 and s+r = 6 (r+s = 36 and 18(s-r) = 36)

Then s = 4 and r = 2

This gives a = 11

Solving further for x1, we get x1 = 3

So the original equation was

x² - 11x + 24 = 0

and the equation was changed to

x² + 11x - 42 = 0

In both cases, the common root is 3

[Guest]

it looks like you guys solved it a little easier then i did. My initial guesses were that this was some sort of trick and the answer was going to be 0,1, or i. I quickly determined these were wrong and then it took me a while but eventually I decided it couldnt be solved by only mathematic manipulation and actually used logic (ingenious isnt it). I found A must be a multiple of 11. and B's digits must be 2 away from each other and from there the rest followed naturally. I found it cool that not only did this question limit the possible equations to 2. But that of the only two possible equations it is the same answer. Anyway as you guys have found.

X² +11x-42

X²+22x-75

x=3

Im confused about your work NOBODY maybe I just didnt follow it right.

hope you enjoyed this. I have another problem im probably going to post later. There are some of the 10 or so problems I found that I havent solved yet. So Im going to try and solve them and then if i think they were fun or interesting Ill post them.

Edited July 29, 2009 by final

[Guest]

Im confused about your work NOBODY maybe I just didnt follow it right.

Hi, I suppose you could follow up to 2n² =9(u-t) Here is 3 unknowns and only one equation, it seems impossible to solve it.

But we know that n,u and t are integers, and both u and t are digits, I mean they're in a range of (1..9).

So get sqrt of our equation: sqrt(2)n=3sqrt(u-t) You can see that sqrt(u-t) should be a multiple of sqrt(2). As u and t are in (1..9), (u-t) can only be 2 or 8.

I (u-t) had been 8, u and t would be 1 and 9 (or 9 and 1). Then, in our equation of (x²-ax+b), b would be 19 or 91. But both are prime numbers and as x₁.x₂=b, this can't be the case.

So (u-t) can only be 2. --> sqrt(2)n=3sqrt(2) -->n=3

I'm also still not sure that this solution is pure mathemathical, or contains some logic or trick? I mean would my math teacher accept this?