Theory of Equations 4 Question 8
9. If $x^{2}+(a-b) x+(1-a-b)=0$ where $a, b \in R$, then find the values of $a$ for which equation has unequal real roots for all values of $b$.
$(2003,4 M)$
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Answer:
Correct Answer: 9. A > 1
Solution:
- Given,
$x^{2}+(a-b) x+(1-a-b)=0$ has real and unequal roots.
$\Rightarrow \quad D>0$
$\Rightarrow \quad(a-b)^{2}-4(1)(1-a-b)>0$
$\Rightarrow a^{2}+b^{2}-2 a b-4+4 a+4 b>0$
Now, to find the values of ’ $a$ ’ for which equation has unequal real roots for all values of $b$.
i.e. Above equation is true for all $b$.
or $b^{2}+b(4-2 a)+\left(a^{2}+4 a-4\right)>0$, is true for all $b$.
$\therefore$ Discriminant, $D<0$
$ \begin{array}{lrl} \Rightarrow & (4-2 a)^{2}-4\left(a^{2}+4 a-4\right) & <0 \\ \Rightarrow & 16-16 a+4 a^{2}-4 a^{2}-16 a+16 & <0 \\ \Rightarrow & -32 a+32<0 \quad \Rightarrow \quad a>1 \end{array} $
