IIT-JEE 2003Question 1
If $x^2+(a-b)x+(1-a-b)=0$ where $a,b\in R$ then find the values of $a$ for which the equation has unequal real roots for all values of $b$.
Quadratic Equation · Nature of Roots and Discriminant · Sign, Range and Inequalities of Quadratic Expressions
Answer$a>1$
Solution
Concept. The discriminant is a quadratic in $b$; it is positive for every $b$ iff its own discriminant is negative. Formulas used. (1) $b^2+Bb+C>0\ \forall b\iff B^2-4C<0$Step 1 - Discriminant of the equation. $$D=(a-b)^2-4(1-a-b)=b^2+(4-2a)b+(a^2+4a-4).$$ Step 2 - Use (1). $$(4-2a)^2-4(a^2+4a-4)=16-16a+4a^2-4a^2-16a+16=32-32a<0.$$ Step 3 - Conclude. $a>1$. Answer: $a>1$