Question

(x+3)(x+7)+1≤(2x-3)²-8(x-5)​

Answer 1

(x+3)(x+7) + 1 ≤ (2x-3)² - 8(x-5)

x² + 10x + 21 + 1 ≤ 4x² - 12x + 9 - 8x + 40

x² + 10x + 22 ≤ 4x² - 20x + 49

0 ≤ 3x² - 30x + 27

3x² - 30x + 27 ≥ 0

x² - 10x + 9 ≥ 0

(x - 1)(x - 9) ≥ 0

x ≤ 1 или x ≥ 9

(-∞, 1]∪[9, +∞).

Answer 2

To solve the inequality $(x+3)(x+7)+1 \leq (2x-3)^2-8(x-5)$, follow these steps:

1. Expand both sides of the inequality: $(x+3)(x+7)+1 \leq (2x-3)^2-8(x-5)$

   Expand the left side: $(x^2 + 10x + 21) + 1 \leq (2x-3)^2 - 8(x-5)$

   Expand the right side: $x^2 + 10x + 22 \leq 4x^2 - 12x + 9 - 8x + 40$

2. Simplify the inequality: Now, we have the inequality in the following form: $(x^2 + 10x + 22) \leq (4x^2 - 12x + 9 - 8x + 40)$

3. Rearrange the inequality by moving all terms to one side to set the inequality to zero: $x^2 + 10x + 22 - 4x^2 + 12x - 9 + 8x - 40 \leq 0$

4. Combine like terms: $-3x^2 + 30x - 27 \leq 0$

5. Divide the entire inequality by -3 to make the leading coefficient of the quadratic term positive, which will help in factoring: $x^2 - 10x + 9 \geq 0$

6. Factor the quadratic expression: $(x-9)(x-1) \geq 0$

7. Find the critical points by setting each factor equal to zero and solving for x: $x - 9 = 0$ => $x = 9$ $x - 1 = 0$ => $x = 1$

8. Create a sign chart to determine the sign of the expression between and outside the critical points:

   \hline \text{Interval} & (x-9) & (x-1) & (x-9)(x-1) & \text{Sign of }(x-9)(x-1) \\ \hline (-\infty, 1) & - & - & + & + \\ (1, 9) & - & + & - & - \\ (9, \infty) & + & + & + & + \\ \hline \end{array} \]

9. Now, analyze the sign chart. The inequality is satisfied when $(x-9)(x-1) \geq 0$, which means when the sign of the expression is either positive or zero. From the sign chart, this occurs in two intervals:

   • $(- \infty, 1)$
   • $(9, \infty)$

10. So, the solution to the inequality is: $-\infty < x \leq 1$ or $x \geq 9$

This is the solution to the given inequality.