Question
(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?
Ask by Moreno Ball. in South Africa
Feb 19,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
To determine the values of \( m \) for which certain quadratic equations have real roots, we analyze each equation's discriminant.
1. **For \( x^{2} - 4x + (2m - 1) = 0 \)**:
- Discriminant \( D = 16 - 8m \geq 0 \) implies \( m \leq 2.5 \).
2. **For \( x^{2} - 8x + 2m = 0 \)**:
- Discriminant \( D = 64 - 8m \geq 0 \) implies \( m \leq 8 \).
3. **For \( x^{2} - 6x - m = 0 \)**:
- Discriminant \( D = 36 + 4m \geq 0 \) implies \( m \geq -9 \).
4. **For \( \frac{x^{2} - 5x + 5}{2x^{2} - x + 1} = \frac{1}{m} \)**:
- Cross-multiplying leads to a quadratic in \( x \) with a discriminant that must be non-negative for real roots.
5. **For \( x^{2} + (k - 2)x - k = 0 \)**:
- The discriminant \( D = (k - 2)^2 + 4k \) is always non-negative, ensuring real roots for all real \( k \).
6. **For \( x^{2} + 4t + 3 = 0 \)**:
- The discriminant \( D = 16 - 12 = 4 \geq 0 \) confirms real roots for all \( t \).
7. **For equal roots in \( x^{2} + 4t + 3 = 0 \)**:
- Set the discriminant to zero: \( 16 - 12 = 0 \), which is always true, meaning equal roots exist for all \( t \).
**Summary**:
- The value of \( m \) must satisfy \( -9 \leq m \leq 2.5 \) for the first two equations to have real roots.
- The equation \( x^{2} + (k - 2)x - k = 0 \) has real roots for all real values of \( k \).
- The equation \( x^{2} + 4t + 3 = 0 \) has real roots for all real values of \( t \).
- The condition for equal roots in the last equation is always satisfied for any real \( t \).
Solution
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!
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Bonus Knowledge
Let's dive into some fun math exploration! For the equation \( x^{2}+(k-2) x-k=0 \) to have real roots, the discriminant must be non-negative. The discriminant \(\Delta\) is given by \( (k-2)^2 - 4(-k) \). Simplifying this leads to \( (k-2)^2 + 4k \). This expression will always be greater than or equal to 0 for all real \( k \), indicating that real roots exist for every \( k \) — a small but mighty math victory! Now let's tackle the second part! The equation \( x^{2}+(t+3)x+1=-2 \) simplifies to \( x^{2}+(t+3)x+3=0 \). For this quadratic to have equal roots, the discriminant must be zero. Thus, \( (t+3)^2 - 4(3) = 0 \). Solving for \( t \) gives us \( t = -3 \pm 2 \sqrt{3} \). These values ensure that the roots won't just exist; they'll be perfectly equal — a thrilling twist in the quadratic world!
