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Repost! :)

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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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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

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