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Let $a$ and $b$ be real numbers such that the quadratic equations $x^2 + ax + b = 0$ and $ax^2 + bx + 1 = 0$ have a root in common. Enter all possible values of $a + b,$ separated by commas.

 
 Jan 14, 2021
 #1
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Not totally sure about this one.....but.....here's what I found

 

Using the  second equation

 

ax^2   +  bx  +  1   =  0

 

ax^2  +  bx  =  -1

 

x ( ax + b)  =   -1

 

ax + b =    -1/x

 

Sub this into the first equation

 

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

 

x^3   -  1   =    0     factor as a difference  of  cubes

 

( x - 1)  (x^2  + x + 1)  =  0

 

Only the first   produces a  single real solution     x   = 1

 

So....using the first equation 

 

(1)^2  + a(1)  +  b   = 0

 

1  +  a +  b  =   0

 

a  +  b   =  -1

 

 

cool cool cool

 
 Jan 14, 2021
edited by CPhill  Jan 14, 2021
 #2
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Hi Cphill, u are completely corrrect, but they ask for multiple values, I suspect in $(x-1)(x^2+x+1)$, we also have to solve for $(x^2+x+1)=0$ to get the other 2 values since it always have two values.

 
 Jan 14, 2021
 #3
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I'm guessing because $r^2+r+1=0$

   comparing:              $ax^2+bx+1=0$

therefore a=1, b=1, therefore resulting in $a+b=2$

 
 Jan 14, 2021

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