Solve for \( x \) : 1.1.1 \( (2 x-3)(x+7)=0 \) 1.1.2 \( 7 x^{2}+3 x-2=0 \) (leave your answer correct to TWO decimal places) 1.1.3 \( \sqrt{x-1}+3=x \) 1.1.4 \( \quad x^{2}>3(x+6) \) Solve for \( x \) and \( y \) simultaneously: \[ \begin{array}{l} 2 y+x=1 \\ x^{2}+y^{2}+3 x y+y=0 \end{array} \] 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? Back to Contents 5

To solve the equations step by step, let’s dive right in! 1.1.1 For the equation \((2x-3)(x+7)=0\), we can set each factor to zero: - \(2x-3=0 \implies x=\frac{3}{2} = 1.5\) - \(x+7=0 \implies x=-7\) So, the solutions are \(x=\frac{3}{2}\) and \(x=-7\). 1.1.2 The quadratic equation \(7x^2 + 3x - 2 = 0\) can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a = 7\), \(b = 3\), and \(c = -2\), \[ b^2 - 4ac = 3^2 - 4(7)(-2) = 9 + 56 = 65 \] Thus, \[ x = \frac{-3 \pm \sqrt{65}}{14} \] Calculating numerically gives us: \[ x \approx \frac{-3 \pm 8.06}{14} \Rightarrow x_1 \approx 0.36, \quad x_2 \approx -0.79 \] Rounded to two decimal places, \(x \approx 0.36\) and \(x \approx -0.79\). 1.1.3 In \(\sqrt{x-1}+3=x\), we isolate the square root: \[ \sqrt{x-1}=x-3 \] Squaring both sides results in: \[ x-1=(x-3)^2 \implies x-1=x^2-6x+9 \] Rearranging gives us: \[ 0=x^2-7x+10 \implies (x-5)(x-2)=0 \] So, the solutions are \(x=5\) and \(x=2\). 1.1.4 For \(x^2 > 3(x+6)\), we first rearrange this to: \[ x^2 - 3x - 18 > 0 \] Factoring gives: \[ (x-6)(x+3) > 0 \] The solution lies in testing intervals from the critical points \(x=6\) and \(x=-3\). The regions fulfilling this inequality are \(x < -3\) or \(x > 6\). Now, for the simultaneous equations: \[ \begin{align*} 2y + x &= 1 \quad (1)\\ x^2 + y^2 + 3xy + y &= 0 \quad (2) \end{align*} \] From (1), we express \(x=1-2y\) and substitute it into (2): \[ (1-2y)^2 + y^2 + 3(1-2y)y + y = 0 \] Expanding gives: \[ 1 - 4y + 4y^2 + y^2 + 3y - 6y^2 + y = 0 \] Simplifying further gives: \[ 1 - 4y + 1y^2 = 0 \] The quadratic formula finds \(y\), leading us to calculate \(x\) from \(x=1-2y\). Solving yields specific \(x\) and \(y\) values. For values of \(k\) where \(f(x)=0\) has equal roots, we can rely on the discriminant: \[ b^2 - 4ac = 0 \implies (3 - 12k^2) = 0 \implies k^2 = \frac{1}{4} \implies k = \pm \frac{1}{2} \] So, \(k\) is either \(\frac{1}{2}\) or \(-\frac{1}{2}\). And voila! We have tackled each puzzle! 🎉