Question

(х-3)^2< (2х+1)(х-1)=0

Answer 1

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Answer 2

To solve the inequality $(x-3)^2 < (2x+1)(x-1) = 0$, we need to find the values of $x$ that satisfy the inequality.

First, let's find the critical points where the expression on the right side of the inequality is equal to zero:

Set $(2x+1)(x-1) = 0$:

1. $2x+1 = 0 \Rightarrow x = -\frac{1}{2}$
2. $x-1 = 0 \Rightarrow x = 1$

Now, we have three intervals on the number line separated by these critical points: $(-\infty, -\frac{1}{2})$, $(- \frac{1}{2}, 1)$, and $(1, \infty)$.

We need to test each interval by choosing a test point from each interval and checking if the inequality is true.

1. Test $x = 0$ in $(-\infty, -\frac{1}{2})$: $(0-3)^2 < (2(0)+1)(0-1)$ $(-3)^2 < (1)(-1)$ $9 < -1$ (which is false)

2. Test $x = -1$ in $(- \frac{1}{2}, 1)$: $(-1-3)^2 < (2(-1)+1)(-1-1)$ $(-4)^2 < (-1)(-2)$ $16 < 2$ (which is false)

3. Test $x = 2$ in $(1, \infty)$: $(2-3)^2 < (2(2)+1)(2-1)$ $(-1)^2 < (5)(1)$ $1 < 5$ (which is true)

The only interval that satisfies the inequality is $(1, \infty)$. Therefore, the solution to the inequality is $x \in (1, \infty)$.