Question

What are the critical points for the inequality x^2 - 4/x^2 - 5x +6 < 0?

Answer 1

Answer:

Critical points are: 2 , -2 and 3.

Step-by-step explanation:

We have been given the expression:

$\frac{x^2-4}{x^2-5x+6}$

We will factorize the given expression:

Using $a^2-b^2=(a+b)(a-b)$

$\frac{(x+2)(x-2)}{x^2-3x-2x+6}$

$\Rightarrow \frac{(x+2)(x-2)}{x(x-3)-2(x-3)}$

$\Rightarrow \frac{(x+2)(x-2)}{(x-2)(x+3)}$

Critical point are those values of an expression when it is equal to zero

Hence, the critical points are: 2,-2 and -3.

Answer 2

Answer:

d. x = –2, x = 2, and x = 3 are critical points.

Step-by-step explanation:

Given : $\frac{x^{2}-4 }{x^{2}-5x+6}$ < 0.

To find :What are the critical points for the inequality.

Solution : We have given that

$\frac{x^{2}-4 }{x^{2}-5x+6}$ < 0.

Using $a^{2}-b^{2} = (a+b)(a-b)$

$\frac{(x-2)(x+2) }{x^{2}-5x+6}$ < 0.

Now, on factoring denominator

$x^{2} -5x+6$

$x^{2} -3x -2x+6$

Taking commom

x( x- 3) -2(x -3)

On grouping (x-3)(x -2)

$\frac{(x-2)(x+2) }{(x-3)(x-2)}$ .

We need to find critical point ,

Critical point on which expression is zero

Then x = 2, -2, 3 are critical point.

Therefore, d. x = –2, x = 2, and x = 3 are critical points.