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Find the number of ordered pairs (a,b) of real numbers such that

 \(a\) is a root of \(x^2+ax+b\)

 \(b\) is a root of \(x^2+ax+b\)

Guest Nov 30, 2018
 #1
avatar+3198 
+1

deleted.

Rom  Nov 30, 2018
edited by Rom  Nov 30, 2018
edited by Rom  Nov 30, 2018
 #2
avatar+92814 
+1

If a is a root, then

 

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

2a^2 + b = 0

b = -2a^2    (1)

 

If b is a root, then

b^2 + ab + b = 0    (2)

 

Sub (1) into (2)

 

(-2a^2)^2 +a(-2a^2) + (-2a^2) = 0    simplify

 

4a^4 -2a^3 - 2a^2 = 0   factor

 

(2a^2) (2a^2 - a - 1) = 0

 

Set both factors to 0 and solve

 

2a^2 = 0        (2a + 1) (a - 1) = 0

a = 0                a = -1/2       a = 1

 

Using (1).....

When a = 0, b = 0     (trivial solution)

When a = -1/2 , b = -1/2

When  a = 1, b = -2

 

Proof

 

x^2 + 0x +0 = 0  .......has the root x = 0.....(a,b) = (0,0)

 

 

x^2 - (1/2)x - 1/2 = 0     multiply through by 2

2x^2  - x^2 - 1  =  0  factor

(2x + 1) (x - 1) = 0

Has the roots  x = -1/2  and x  = 1 

So  .....(a,b) = (-1/2, -1/2)

 

x^2 -2x + 1 = 0      factor

(x -1)  (x -1) = 0.......so x = 1 is a root......(a, b) = (1, -2)

 

So......

 

(a, b)  = (0, 0)    (trivial)

(a, b) = (-1/2, -1/2)

(a, b) = (1, -2)

 

cool cool cool

CPhill  Nov 30, 2018
edited by CPhill  Nov 30, 2018

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