Question 32196
Hello!
Try plugging x = 1 into the equation. You get:

{{{a*1 + b*1 + a =0 }}}
{{{2a + b = 0}}}
{{{b = -2a}}}

However, b is odd, but 2a is even (two multiplied by any number yields an even number). Therefore, x=1 can't be a solution to the equation.

When plugging x = -1, we get:

{{{a*1 +b*(-1) + a = 0}}}
{{{a - b + a = 0}}}
{{{b = 2a}}}

Using the same reasoning, we conclude that x=-1 can't be a solution of your equation.


I hope this helps!
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