Find all possible values of 'a' for which exactly one root of the quadratic equation $x^{2} - (a + 1) x + 2 a = 0$ lie in the interval $(0, 3)$

Open in App

Solution

$x^{2} - (a + 1) x + 29 = 0$ $1 (0, 3)$ $p = 0, q = 3$

For exactly one root

$f (p) . f (a) < 0$

$\Rightarrow f (a) \times f (3) < 0$

$\Rightarrow 2 a \times [9 - {(a + 1)}^{3} + 2 a] < 0$

$\Rightarrow 2 a (9 - 3 a - 3 + 2 a) < 0$

$\Rightarrow 2 a (6 - a) < 0$

$a \in (- \infty, 0) \cup (6, \infty)$

Suggest Corrections

Similar questions

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

(0, 3)

a

a

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

(0, 3)

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

(0, 3) (0) (3) < 0

\Rightarrow 2 a (9 - 3 (a + 1) + 2 a) < 0

\Rightarrow 2 a (- a + 6) < 0

\Rightarrow a (a - 6) > 0

\Rightarrow a < 0 o r a > 6

x^{2} + a x + 2 = 0

(0, 3)

a

x^{2} - (m - 3) x + m = 0 (m \in R)

m

(1, 2)

Join BYJU'S Learning Program