Question

Let s be a set of values of 'a' for which 2 lie between the roots of quadratic equation x2+(a+2)x-(a+3)=0 then s is given by.

Answer 1

Answer:

$x = - 5$

and

$x = 1$

Step-by-step explanation:

given that a=2

so,

${x}^{2} + (a + 2)x - (a + 3)$

now,

${x }^{2} + (2 + 2)x - (2 + 3)x = 0$

${x}^{2} + 4x - 5 = 0$

${x}^{2} + 5x - x - 5 = 0$

$x(x + 5) - 1(x + 5) = 0$

$(x + 5)(x - 1) = 0$

$x + 5 = 0$

$x = - 5$

$x - 1 = 0$

$x = 1$