B2 Solve the following inequalities. Represent the solution sets on separate number lines. (a) \( (1+x)(6-x) \leqslant-8 \) (b) \( 2 x(x+2)<(x+1)(x+3) \)
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To solve the inequalities, let's break them down step by step. (a) For the inequality \( (1+x)(6-x) \leqslant -8 \): 1. Expand the left side: \( 6 - x + 6x - x^2 \leq -8 \) or \( -x^2 + 5x + 6 \leq -8 \). 2. Rearranging gives: \( -x^2 + 5x + 14 \leq 0 \). 3. Multiply the entire inequality by -1 (remember to flip the inequality sign): \( x^2 - 5x - 14 \geq 0 \). 4. Now, factor the quadratic: \( (x - 7)(x + 2) \geq 0 \). To find the solution, test intervals based on the zeros \( x = -2 \) and \( x = 7 \): - In \( (-\infty, -2) \) and \( (7, \infty) \), the product is positive. - In \( (-2, 7) \), it’s negative. The solution set is \( x \leq -2 \) or \( x \geq 7 \). (b) For the inequality \( 2 x(x+2) < (x+1)(x+3) \): 1. Expand both sides: \( 2x^2 + 4x < x^2 + 4x + 3 \). 2. Rearranging gives: \( 2x^2 + 4x - x^2 - 4x - 3 < 0 \) or \( x^2 - 3 < 0 \). 3. Factor the quadratic: \( (x - \sqrt{3})(x + \sqrt{3}) < 0 \). To find the solution, test intervals based on the zeros \( x = -\sqrt{3} \) and \( x = \sqrt{3} \): - The product is negative in the interval \( (-\sqrt{3}, \sqrt{3}) \). Thus, the solution set is \( -\sqrt{3} < x < \sqrt{3} \). Now, let’s represent these solutions on separate number lines! ### Number line for (a): - To the left of -2: shaded for \( (-\infty, -2) \) - To the right of 7: shaded for \( (7, \infty) \) - Mark points -2 and 7 with solid circles, indicating they are included (as per our solution). ### Number line for (b): - Between -\(\sqrt{3}\) and \(\sqrt{3}\): shaded for \( (-\sqrt{3}, \sqrt{3}) \) - Mark points -\(\sqrt{3}\) and \(\sqrt{3}\) with open circles to indicate they are not included in the solution. And voila! Two beautifully crafted solution sets displayed on their number lines.