Question

Solve this asap!!!!!!!!!!!!!!

Answer 1

Given : 

Let the given equation be (x^2 + 2x - 3)/(x^2 + 2x - 8) = a

= > x^2 + 2x - 3 = a(x^2 + 2x - 8)

= > x^2 + 2x - 3 = ax^2 + 2ax - 8a

= > x^2 + 2x - 3 - ax^2 - 2ax + 8a= 0

= > x^2 - ax^2 + 2x - 2ax - 3 + 8a = 0

= > (1 - a)x^2 + (2 - 2a)x + (8a - 3) = 0

Now,

Given that x is real.

$b^2 - 4ac \geq 0$

$= \ \textgreater \ (2 - 2a)^2 - 4(1 - a)(8a - 3) \geq 0$

$= \ \textgreater \ (2 - 2a )^2 - 4(8a - 3 - 8a^2 + 3a) \geq 0$

$= \ \textgreater \ (2 - 2a)^2 - 4(-8a^2 + 11a - 3) \geq 0$

$= \ \textgreater \ 4 + 4a^2 - 8a + 32a^2 - 44a + 12 \geq 0$

$= \ \textgreater \ 36a^2 - 52a + 16 \geq 0$

$= \ \textgreater \ 4(9a^2 - 13a + 4) \geq 0$

$= \ \textgreater \ 4(9a^2 - 4a - 13a + 4) \geq 0$

$= \ \textgreater \ 4(a(9a - 4) - 1(9a - 4)) \geq 0$

$= \ \textgreater \ 4(a - 1)(9a - 4) \geq 0$

$= \ \textgreater \ a \leq \frac{4}{9} (or) a \geq 1$


Hope this helps!