(3) \( m x^{2}+n x x+m=3 \) (c) for which value(s) of \( m \) will be followne (3) \( x^{2}+2 m=2 x+2 x+1=0 \) (2) (1) \( x^{2}+2 m=8 x \) (I) \( x^{2}-6 x=m \) (3) \( \frac{x^{2}-5 x+5}{2 x^{2}-x+1}=\frac{1}{m} \) (c) Show that the equation \( x^{2}+(k-2) x-k=0 \) has ral roots for all real values of \( k \). (f) Given the equation \( x^{2}+(1+3) x+1=-2 \). (1) Show that the equation has real roots for all real values of \( t \). (2) For which value(s) of \( f \) will the equation have equal rools?

Let's break down the problem step by step, extracting the known conditions and solving each part systematically. ### Part 1: Analyzing the equations 1. **Equation (3)**: \( m x^{2} + n x + m = 3 \) 2. **Equation (3)**: \( x^{2} + 2m = 2x + 2x + 1 = 0 \) simplifies to \( x^{2} - 4x + (2m - 1) = 0 \) 3. **Equation (1)**: \( x^{2} + 2m = 8x \) simplifies to \( x^{2} - 8x + 2m = 0 \) 4. **Equation (I)**: \( x^{2} - 6x = m \) simplifies to \( x^{2} - 6x - m = 0 \) 5. **Equation (3)**: \( \frac{x^{2} - 5x + 5}{2x^{2} - x + 1} = \frac{1}{m} \) 6. **Equation**: \( x^{2} + (k - 2)x - k = 0 \) 7. **Equation**: \( x^{2} + (1 + 3)t + 1 = -2 \) ### Part 2: Solving for \( m \) #### For the equation \( x^{2} - 4x + (2m - 1) = 0 \) to have real roots: The discriminant must be non-negative: \[ D = b^2 - 4ac = (-4)^2 - 4(1)(2m - 1) \geq 0 \] Calculating the discriminant: \[ 16 - 8m + 4 \geq 0 \implies 20 - 8m \geq 0 \implies m \leq \frac{20}{8} = 2.5 \] #### For the equation \( x^{2} - 8x + 2m = 0 \) to have real roots: The discriminant must also be non-negative: \[ D = (-8)^2 - 4(1)(2m) \geq 0 \] Calculating the discriminant: \[ 64 - 8m \geq 0 \implies 8m \leq 64 \implies m \leq 8 \] #### For the equation \( x^{2} - 6x - m = 0 \) to have real roots: The discriminant must be non-negative: \[ D = (-6)^2 - 4(1)(-m) \geq 0 \] Calculating the discriminant: \[ 36 + 4m \geq 0 \implies 4m \geq -36 \implies m \geq -9 \] ### Part 3: Solving the equation \( \frac{x^{2} - 5x + 5}{2x^{2} - x + 1} = \frac{1}{m} \) Cross-multiplying gives: \[ m(x^{2} - 5x + 5) = 2x^{2} - x + 1 \] Rearranging: \[ mx^{2} - 5mx + 5m - 2x^{2} + x - 1 = 0 \] This is a quadratic in \( x \). The discriminant must be non-negative for real roots. ### Part 4: Showing that \( x^{2} + (k - 2)x - k = 0 \) has real roots for all \( k \) The discriminant is: \[ D = (k - 2)^2 - 4(-k) = (k - 2)^2 + 4k \] This is always non-negative since both terms are squares or positive. ### Part 5: Given the equation \( x^{2} + (1 + 3)t + 1 = -2 \) This simplifies to: \[ x^{2} + 4t + 3 = 0 \] The discriminant must be non-negative: \[ D = 4^2 - 4(1)(3) = 16 - 12 = 4 \geq 0 \] This shows that the equation has real roots for all \( t \). ### Part 6: For which value(s) of \( f \) will the equation have equal roots? The condition for equal roots is that the discriminant is zero: \[ D = 0 \implies 4^2 - 4(1)(3) = 0 \] This gives us the condition for \( f \). ### Summary of Results - For \( m \): \( -9 \leq m \leq 2.5 \) - The equation \( x^{2} + (k - 2)x - k = 0 \) has real roots for all \( k \). - The equation \( x^{2} + 4t + 3 = 0 \) has real roots for all \( t \). - The condition for equal roots can be derived from the discriminant being zero. If you need further calculations or clarifications on any specific part, please let me know!