Question

if x is real, then the range of x^2-2x=9/x^2+2x+9 is

Answer 1

Answer:

Step-by-step explanation:

Given  

if x is real, then the range of x^2-2x=9/x^2+2x+9 is

Let y = x^2 – 2x – 9 / x^2 + 2 x + 9

y(x^2 + 2x + 9) = (x^2 – 2x – 9)

yx^2 + 2xy + 9y = x^2 – 2x – 9

yx^2 – x^2 + 2xy + 2x + 9y + 9 = 0

(y – 1 ) x^2 + x (2y + 2) + 9y + 9 = 0

We know that b^2 – 4ac > = 0

(2y + 2)^2 – 4(y – 1)(9y + 9) >= 0

4y^2 + 4 + 8y – 4(9y^2 – 9y + 9y – 9)

4y^2 + 4 + 8y – 36y^2 + 36 > = 0

32y^2 – 8y – 40 <=0

4y^2 – y – 5 <=0

4y^2 + 4y - 5y – 5<=0

4y(y + 1) – 5(y + 1) < = 0

(4y – 5)(y + 1) < = 0

Now y = 5/4, - 1

So range will be y ∈  (5/4, -1)