QUESTION I DEE KON 170 1.1 Solve for \( x \). \( 1.11 \quad(2 x-3)(x+7)=0 \) \( 1.127 x^{2}+3 x-2=0 \) (leave your axaser correct to TWO decimal places) \( 1.13 \sqrt{x-1}+3=x \) \( 114 \quad x^{2}>3(x+6) \) 12 Solve for \( x \) and \( y \) simulianoously: \[ \begin{array}{l} 2 y+x=1 \\ x^{2}+y^{2}+3 x y+y=0 \end{array} \] 13 If \( f(x)=0 \) has roo's \( x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} \), for which values of \( k \) will the rcots be equal?

Let's solve each part of the problem step by step. ### 1.11 Solve for \( x \) in \( (2x-3)(x+7)=0 \) To solve this equation, we set each factor equal to zero: 1. \( 2x - 3 = 0 \) 2. \( x + 7 = 0 \) **Step 1: Solve \( 2x - 3 = 0 \)** \[ 2x = 3 \implies x = \frac{3}{2} \] **Step 2: Solve \( x + 7 = 0 \)** \[ x = -7 \] Thus, the solutions are \( x = \frac{3}{2} \) and \( x = -7 \). ### 1.12 Solve for \( x \) in \( x^{2}+3x-2=0 \) We will use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 3, c = -2 \). **Step 1: Calculate the discriminant** \[ b^2 - 4ac = 3^2 - 4(1)(-2) = 9 + 8 = 17 \] **Step 2: Apply the quadratic formula** \[ x = \frac{-3 \pm \sqrt{17}}{2} \] Calculating the two roots: 1. \( x_1 = \frac{-3 + \sqrt{17}}{2} \) 2. \( x_2 = \frac{-3 - \sqrt{17}}{2} \) Now, let's calculate these values to two decimal places. ### 1.13 Solve for \( x \) in \( \sqrt{x-1}+3=x \) **Step 1: Isolate the square root** \[ \sqrt{x-1} = x - 3 \] **Step 2: Square both sides** \[ x - 1 = (x - 3)^2 \] **Step 3: Expand and rearrange** \[ x - 1 = x^2 - 6x + 9 \] \[ 0 = x^2 - 7x + 10 \] **Step 4: Factor the quadratic** \[ 0 = (x - 5)(x - 2) \] Thus, \( x = 5 \) or \( x = 2 \). **Step 5: Check for extraneous solutions** 1. For \( x = 5 \): \[ \sqrt{5-1} + 3 = 2 + 3 = 5 \quad \text{(valid)} \] 2. For \( x = 2 \): \[ \sqrt{2-1} + 3 = 1 + 3 = 4 \quad \text{(not valid)} \] Thus, the only solution is \( x = 5 \). ### 1.14 Solve the inequality \( x^{2}>3(x+6) \) **Step 1: Rearrange the inequality** \[ x^2 - 3x - 18 > 0 \] **Step 2: Factor the quadratic** \[ (x - 6)(x + 3) > 0 \] **Step 3: Determine the intervals** The critical points are \( x = 6 \) and \( x = -3 \). We test intervals: 1. \( (-\infty, -3) \): Choose \( x = -4 \) → \( (-)(-) > 0 \) (valid) 2. \( (-3, 6) \): Choose \( x = 0 \) → \( (-)(+) < 0 \) (not valid) 3. \( (6, \infty) \): Choose \( x = 7 \) → \( (+)(+) > 0 \) (valid) Thus, the solution is \( x < -3 \) or \( x > 6 \). ### 1.12 Solve for \( x \) and \( y \) simultaneously: \[ \begin{array}{l} 2y + x = 1 \\ x^2 + y^2 + 3xy + y = 0 \end{array} \] **Step 1: Solve the first equation for \( x \)** \[ x = 1 - 2y \] **Step 2: Substitute into the second equation** \[ (1 - 2y)^2 + y^2 + 3(1 - 2y)y + y = 0 \] Expanding this: \[ (1 - 4y + 4y^2) + y^2 + (3y - 6y^2) + y = 0 \] \[ 1 - 4y + 4y^2 + y^2 + 3y - 6y^2 + y = 0 \] \[ 1 - 4y + 4y^2 + 3y - 6y^2 + y = 0 \] \[ 1 - 4y + 3y - 2y^2 = 0 \] \[ -2y^2 + 0y + 1 = 0 \implies 2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \] **Step 3: Find corresponding \( x \) values** For \( y = \frac{\sqrt{2}}{2} \): \[ x = 1 - 2\left(\frac{\sqrt{2}}{2}\right) = 1 - \sqrt{2} \] For \( y = -\frac{\sqrt{2}}{2} \): \[ x = 1 - 2\left(-\frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2} \] Thus, the solutions are: 1. \( (1 - \sqrt{2}, \frac{\sqrt{2}}{2}) \) 2. \( (1 + \sqrt{2}, -\frac{\sqrt{2}}{2}) \) ### 1.13 If \( f(x)=0 \) has roots \( x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} \), for which values of \( k \) will the roots be equal? For the roots to be equal, the discriminant must be zero: \[ 3 - 12k^2 = 0 \] **Step 1: Solve for \( k \)** \[ 12k^2 = 3 \implies k^2 =