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!!!
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