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Find all the values of $a$ for which the equations \begin{align*} x^2 + ax + 1 &= 0, \\ x^2  - x - a &= 0 \end{align*}have a common real root. Enter all the possible values, separated by commas.

 

what I did:

 

I set the equations equal getting

 

$x^2+ax+1=x^2-x-a$

simplifying

$ax+x+a+1=0$

factor we get 

$(a+1)(x+1)=0$

 

 

IM STUCK!!!

 Jan 25, 2021
 #1
avatar+116126 
+1

Complete the square  on the  first equation                   Complete the square  on the  second equation 

 

x^2 + ax + a^2/4   =   -1 + a^2/4                                        x^2 -  x  + 1/4  =  a + 1/4

 

(x + a/2)^2  =  (a^2  - 4)  /4                                               ( x - 1/2)^2  = ( 4a + 1) / 4

 

                                                    Take  both roots  

 

( x + a/2)  =  ± √(a^2 - 4)  /2                                             ( x  -1/2)   = ± √(a + 1/4) /2

 

x = ± √(a^2  - 4) / 2   - a/2                                                  x   =  ± √(4a + 1) / 2  + 1/2

 

Note  that   when  a   = 2      

 

x  =   √ (2^2 - 4) / 2   -  2/2  =  -1                                      x  =  -  √  ( 9)/ 2  + 1/2 = -3/2 + 1/2  =  -1

 

So

x^2  + 2x  + 1  =  0      has the double  root of  -1

And

x^2  - x - 2  = 0           has the  roots  2  and  -1 

 

So....they share  the  root  of  -1      when  a  =  2

 

cool cool cool

 Jan 25, 2021
 #2
avatar
0

Thx bro! Is that the only value

 

anyways I got a=-1, 2 by using the quadratic formula on both, setting them equal and then solving for a, then changing the plus minus signs for all possibilities of signs,

 Jan 25, 2021
 #3
avatar
0

But a=-1 is incorrect btw

 Jan 25, 2021
 #4
avatar+116126 
0

"a"  cannot be   -1     because   x^2 -  x + 1    = 0    has no  real roots  (the discriminant is <  0  )

 

cool cool cool

 Jan 25, 2021

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