1.1 Solve for \( x \) : 1.1.1 \( x(x+1)=0 \) 1.1.2 \( 3 x^{2}-5 x+1=0 \) (correct to TWO decimal places) (4) 1.1.3 \( 2\left(x^{2}-3 x\right)>-4 \) (4) 1.2 Solve for \( x \) and \( y \) simultaneously if: \[ x=4-y \text { and } 2^{x} \times 2^{y}=2^{x^{2}+3} \] 1.3 For which value of \( r \) will \( 2 x^{2}-2 x=r \) have equal roots? (4)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x\left(x+1\right)=0\) - step1: Separate into possible cases: \(\begin{align}&x=0\\&x+1=0\end{align}\) - step2: Solve the equation: \(\begin{align}&x=0\\&x=-1\end{align}\) - step3: Rewrite: \(x_{1}=-1,x_{2}=0\) Solve the equation \( 3 x^{2}-5 x+1=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(3x^{2}-5x+1=0\) - step1: Solve using the quadratic formula: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 3}}{2\times 3}\) - step2: Simplify the expression: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 3}}{6}\) - step3: Simplify the expression: \(x=\frac{5\pm \sqrt{13}}{6}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{5+\sqrt{13}}{6}\\&x=\frac{5-\sqrt{13}}{6}\end{align}\) - step5: Rewrite: \(x_{1}=\frac{5-\sqrt{13}}{6},x_{2}=\frac{5+\sqrt{13}}{6}\) Solve the equation \( 2\left(x^{2}-3 x\right)>-4 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(2\left(x^{2}-3x\right)>-4\) - step1: Move the expression to the left side: \(2\left(x^{2}-3x\right)-\left(-4\right)>0\) - step2: Subtract the terms: \(2x^{2}-6x+4>0\) - step3: Rewrite the expression: \(2x^{2}-6x+4=0\) - step4: Factor the expression: \(2\left(x-2\right)\left(x-1\right)=0\) - step5: Divide the terms: \(\left(x-2\right)\left(x-1\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-2=0\\&x-1=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=2\\&x=1\end{align}\) - step8: Determine the test intervals: \(\begin{align}&x<1\\&12\end{align}\) - step9: Choose a value: \(\begin{align}&x_{1}=0\\&x_{2}=\frac{3}{2}\\&x_{3}=3\end{align}\) - step10: Test the chosen value: \(\begin{align}&x<1\textrm{ }\textrm{is the solution}\\&12\textrm{ }\textrm{is the solution}\end{align}\) - step11: The final solution is \(x \in \left(-\infty,1\right)\cup \left(2,+\infty\right):\) \(x \in \left(-\infty,1\right)\cup \left(2,+\infty\right)\) Solve the equation \( 2 x^{2}-2 x=r \). Solve the equation by following steps: - step0: Solve for \(x\): \(2x^{2}-2x=r\) - step1: Move the expression to the left side: \(2x^{2}-2x-r=0\) - step2: Solve using the quadratic formula: \(x=\frac{2\pm \sqrt{\left(-2\right)^{2}-4\times 2\left(-r\right)}}{2\times 2}\) - step3: Simplify the expression: \(x=\frac{2\pm \sqrt{\left(-2\right)^{2}-4\times 2\left(-r\right)}}{4}\) - step4: Simplify the expression: \(x=\frac{2\pm \sqrt{4+8r}}{4}\) - step5: Simplify the expression: \(x=\frac{2\pm 2\sqrt{1+2r}}{4}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{2+2\sqrt{1+2r}}{4}\\&x=\frac{2-2\sqrt{1+2r}}{4}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{1+\sqrt{1+2r}}{2}\\&x=\frac{2-2\sqrt{1+2r}}{4}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=\frac{1+\sqrt{1+2r}}{2}\\&x=\frac{1-\sqrt{1+2r}}{2}\end{align}\) Solve the system of equations \( x=4-y; 2^{x} \times 2^{y}=2^{x^{2}+3} \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x=4-y\\2^{x}\times 2^{y}=2^{x^{2}+3}\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x=4-y\\x+y=x^{2}+3\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4-y+y=\left(4-y\right)^{2}+3\) - step3: Simplify: \(4=19-8y+y^{2}\) - step4: Swap the sides: \(19-8y+y^{2}=4\) - step5: Move the expression to the left side: \(19-8y+y^{2}-4=0\) - step6: Subtract the numbers: \(15-8y+y^{2}=0\) - step7: Factor the expression: \(\left(5-y\right)\left(3-y\right)=0\) - step8: Separate into possible cases: \(\begin{align}&5-y=0\\&3-y=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=5\\&y=3\end{align}\) - step10: Calculate: \(y=5\cup y=3\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=4-y\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=4-y\\y=3\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-1,5\right)\cup \left(x,y\right) = \left(1,3\right)\) Let's solve each part step by step. ### 1.1 Solve for \( x \) #### 1.1.1 \( x(x+1)=0 \) To solve the equation \( x(x+1)=0 \), we can set each factor to zero: 1. \( x = 0 \) 2. \( x + 1 = 0 \) which gives \( x = -1 \) Thus, the solutions are: \[ x_{1} = -1, \quad x_{2} = 0 \] #### 1.1.2 \( 3 x^{2}-5 x+1=0 \) We will use the quadratic formula to solve this equation, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = -5 \), and \( c = 1 \). Calculating the discriminant: \[ b^2 - 4ac = (-5)^2 - 4 \cdot 3 \cdot 1 = 25 - 12 = 13 \] Now applying the quadratic formula: \[ x = \frac{5 \pm \sqrt{13}}{6} \] Calculating the two roots: 1. \( x_{1} = \frac{5 - \sqrt{13}}{6} \approx 0.23 \) 2. \( x_{2} = \frac{5 + \sqrt{13}}{6} \approx 1.43 \) Thus, the solutions rounded to two decimal places are: \[ x_{1} \approx 0.23, \quad x_{2} \approx 1.43 \] #### 1.1.3 \( 2\left(x^{2}-3 x\right)>-4 \) First, we simplify the inequality: \[ 2x^{2} - 6x + 4 > 0 \] Dividing through by 2: \[ x^{2} - 3x + 2 > 0 \] Factoring the quadratic: \[ (x-1)(x-2) > 0 \] To find the intervals where this inequality holds, we analyze the critical points \( x = 1 \) and \( x = 2 \). The test intervals are: - \( (-\infty, 1) \) - \( (1, 2) \) - \( (2, +\infty) \) Testing these intervals: - For \( x < 1 \) (e.g., \( x = 0 \)): \( (0-1)(0-2) > 0 \) is true. - For \( 1 < x < 2 \) (e.g., \( x = 1.5 \)): \( (1.5-1)(1.5-2) < 0 \) is false. - For \( x > 2 \) (e.g., \( x = 3 \)): \( (3-1)(3-2) > 0 \) is true. Thus, the solution is: \[ x \in (-\infty, 1) \cup (2, +\infty) \] ### 1.2 Solve for \( x \) and \( y \) simultaneously Given the equations: 1. \( x = 4 - y \) 2. \( 2^{x} \times 2^{y} = 2^{x^{2}+3} \) Substituting \( x \) from the first equation into the second: \[ 2^{4-y} \times 2^{y} = 2^{(4-y)^{2}+3} \] This simplifies to: \[ 2^{4} = 2^{(4-y)^{2}+3} \] Thus, we have: \[ 4 = (4-y)^{2} + 3 \] Subtracting 3 from both sides: \[ 1 = (4-y)^{2} \] Taking the square root: \[ 4 - y = 1 \quad \text{or} \quad 4 - y = -1 \] This gives: 1. \( y = 3 \) leading to \( x = 1 \) 2. \( y = 5 \) leading to \( x = -1 \) Thus, the solutions are: \[ (x,y) = (-1,5) \quad \text{or} \quad (x,y) = (1,3) \] ### 1.3 For which value of \( r \) will \( 2 x^{2}-2 x=r \) have equal roots? For a quadratic equation to have equal roots, the discriminant must be zero. The equation can be rewritten as: \[ 2x^{2} - 2x - r = 0 \] The discriminant \( D \) is given by: \[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot (-r) = 4 + 8r \] Setting the discriminant to zero for equal roots: \[ 4 + 8r = 0 \] Solving for \( r \): \[ 8r = -4 \quad \Rightarrow \quad r = -\frac{1}{2} \] Thus, the value of \( r \) for which the equation has equal roots is: \[ r = -\frac{1}{2} \]