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B2 Solve the following inequalities. Represent the solution sets on separate number lines. (a) \( (1+x)(6-x) \leqslant-8 \) (b) \( 2 x(x+2)<(x+1)(x+3) \)

Ask by Summers Knight. in Singapore
Mar 22,2025

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Answer

**(a)** The solution is \( x \leq -2 \) or \( x \geq 7 \). **(b)** The solution is \( -\sqrt{3} < x < \sqrt{3} \).

Solution

**(a)** We start with the inequality \[ (1+x)(6-x) \leq -8. \] *Step 1: Expand the left-hand side.* \[ (1+x)(6-x)=6+5x-x^2. \] *Step 2: Rewrite the inequality.* \[ 6+5x-x^2 \le -8. \] *Step 3: Bring all terms to one side.* \[ -x^2+5x+6+8 \le 0 \quad \Longrightarrow \quad -x^2+5x+14 \le 0. \] *Step 4: Multiply both sides by \(-1\) (remember to reverse the inequality sign).* \[ x^2-5x-14 \ge 0. \] *Step 5: Factor the quadratic expression (or use the quadratic formula).* Calculate the discriminant: \[ \Delta = (-5)^2-4(1)(-14)=25+56=81. \] Find the roots: \[ x=\frac{5\pm\sqrt{81}}{2}=\frac{5\pm9}{2}. \] Thus, the roots are \[ x=-2 \quad \text{and} \quad x=7. \] *Step 6: Analyze the sign of the quadratic \(x^2-5x-14=(x-7)(x+2)\).* Since the quadratic has a positive leading coefficient, it is nonnegative when \[ x\leq -2 \quad \text{or} \quad x\geq 7. \] *Solution for (a):* \[ x \in (-\infty, -2] \cup [7, \infty). \] *Number line for (a):* Place a closed circle at \(-2\) and at \(7\) with shading to the left of \(-2\) and to the right of \(7\). --- **(b)** We start with the inequality \[ 2x(x+2) < (x+1)(x+3). \] *Step 1: Expand both sides.* Left-hand side: \[ 2x(x+2)=2x^2+4x. \] Right-hand side: \[ (x+1)(x+3)=x^2+4x+3. \] *Step 2: Set up the inequality.* \[ 2x^2+4x < x^2+4x+3. \] *Step 3: Subtract \(x^2+4x\) from both sides.* \[ 2x^2+4x - (x^2+4x) <3 \quad \Longrightarrow \quad x^2<3. \] *Step 4: Solve \(x^2<3\).* Taking square roots and considering both positive and negative solutions gives \[ -\sqrt{3} < x < \sqrt{3}. \] *Solution for (b):* \[ x \in (-\sqrt{3}, \sqrt{3}). \] *Number line for (b):* Draw an open circle at \(-\sqrt{3}\) and at \(\sqrt{3}\) with shading in the interval between them.

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The Deep Dive

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.

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