Consider the equation given below. $x^{2} + (a - b) x + (1 - a - b) = 0, b \in R$ .Find the condition on $^{'} a^{'}$ for which both roots of the equation are real and unequal $\forall b \in R$ .
Roots are imaginary $\forall b ϵ R$

Open in App

Solution

Consider the given equation.
$x^{2} + (a - b) x + (1 - a - b) = 0$

If both the roots are to be real and unequal. we know that
$b^{2} - 4 a c > 0$
$(a - b)^{2} - 4 \times (1 - a - b) > 0$
$a^{2} + b^{2} - 2 a b - 4 + 4 a + 4 b > 0$
$b^{2} - (2 a - 4) b + a^{2} + 4 a - 4 > 0$

Since, the above equation is always greater than $0$ , it will not have any real roots. Thus, we have
$(2 a - 4)^{2} - 4 (a^{2} + 4 a - 4) < 0$
$4 a^{2} - 16 a + 16 - 4 a^{2} - 16 a + 16 < 0$
$- 32 a < - 32$
$a < 1$

Thus, if the equation has to always have real and unequal roots $\forall b \in R$ , $a < 1$ .

Suggest Corrections

Similar questions

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

a, b \in R .

a

\frac{11 x^{2} + 12 x + 6}{x^{2} + 4 x + 2}

- 5

3

x^{2} + (a - b) x + (1 - a - b) = 0, a, b ϵ R

x^{2} - 3 x + 2 > 0

x^{2} - 3 x - 4 \leq 0

a, b, c \in R^{+}

(x - b) (x - c) + (x - a) (x - c) + (x - a) (x - b) = 0

x^{2} - 2 c x + a b = 0

x^{2} - 2 (a + b) x + a^{2} + b^{2} + 2 c^{2} = 0

2 x^{2} - 6 x + 3 = 0

Join BYJU'S Learning Program