View Full Version : Totally awesome problem!

bigduke

10-25-2004, 11:23 AM

If 3 + sqrt(2) satisfies the equation x^2 + ax + b = 0 where a and b are rational, then a and b are respectively:

A) 7, -6

B) -6, -8

C) 8, -7

D) -7, 8

E) -6, 7

I solved this long time back and got it correct, wonder what I'm missing out this time.

Ans. E

Hey bigdude here is a method to solve it quickly (method of undetermined coefficients)

For easyness I will write sqrt(2) as s.

Plug in 3+s and take:

(3+s)^2 + a(3+s) + b = 0

9 + 2 + 6s + a(3+s) + b = 0

a(3+s) + b = -11 - 6s (try to make LHS and RHS have similar form)

a(3+s) + b = 7 - 18 -6s

a(3+s) + b = -6(3+s) + 7

thus LHS=RHS ----------------> a=-6 and b=7.

You are not missing anything by plagging the answers in and trying to find something true. Where did you find this question? I think it is a little bit difficult for the GRE.

Dingus

10-25-2004, 03:39 PM

Another way here:

-----------------------------------------------------------------------------

Any quadratic equation can be expressed as:

ax^2 + bx + c = 0

where the sum of roots is --> -b/a

and the product of roots --> c/a

Hence the equation can be written as:

x^2 -(sum of the roots)x + (product of roots) =0

--> x^2 -Sx +P =0

------------------------------------------------------------------------------

Now another fact:

If one of the roots is irrational another root is also irrational, but its conjugate.

For eg:

One of the roots--> 3+ sqrt2

Then the other root -->3 - sqrt2

Similar roots are --> sqrt2 and - sqrt2, sqrt5 + sqrt3 and sqrt5 - sqrt3 and so on.

To recap, irrational roots *always* occur in pairs and are conjugates of each other.

--------------------------------------------------------------------------------

In this problem the roots are 3 + sqrt2 and 3 - sqrt2

Hence the equation is:

x^2 - (6)x +(7) =0

(using fact1--> x^2 - Sx +P= 0)

Compare and you get --> a=-6 and b=7

bigduke

10-25-2004, 03:40 PM

Well I have a set of assignments from my course I took for GRE prep last year.

They probably believe in the "light bulb" training method :D

lmtuan

10-26-2004, 03:12 AM

Replace x = 3 + sqrt(2) into the equation x^2 + ax + b = 0, we have

[ 11 + 6*sqrt(2) ] + a [ 3 + sqrt(2) ] + b = 0

[ 6 + a]*sqrt(2) = -b - 11 - 3*a

Since a and b are rational, sqrt(2) is irrational . Then, by identifying two side, we have

6 + a = 0 and b + 11 + 3*a = 0. Solve two equations, we receive that a=-6 and b=7.

ANSWER: E

A) 7, -6

B) -6, -8

C) 8, -7

D) -7, 8

E) -6, 7

calm_J

11-06-2004, 11:49 AM

hey both techniques are good but i believe in what i c.

which is the application of the eqn. ax^2+bx+c = 0 and use its rule to find 2 eqns to solve..... was not quite good as i expected and said in a previous post

c u

renriqu1

11-06-2004, 02:48 PM

Hey guys, here's a really fast way to look at it. Use the quadratic equation:

For the example here: x^2 + ax + b, x = 3 + sqrt(2).

x is therefore [-a +/- sqrt(a^2 - 4*1*b)]/(2*1) Take note of the brackets and parantheses

Then do it by parts...first solve for a.

-a/2 = 3 -----> a = -6

Now plug this into the second part..

sqrt((-6)^2 - 4b)]/2 = sqrt(2)

sqrt(36 - 4b)] = 2 sqrt(2) = (sqrt(4))*(sqrt(2)) = ((sqrt(8))

Now drop the square roots...

36 -4b = 8

4b = 28 -----> b = 7