Question 995840
The first part is wrong.  That cannot be justified.  An example was already given a few minutes ago.


Use the discriminant for the second question.
{{{x^2+bx+1=0}}} --- given equation;
{{{b^2-4*1*1}}} --- the discriminant, using a=1 and c=1;
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{{{b^2-4>=0}}} --- discriminant must be non-zero for the original quadratic equation to have real solutions.
{{{b^2=4}}}
{{{highlight(b=0+- 2)}}}