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Sagot :
To find the values of [tex]\(k\)[/tex] for which the roots of the quadratic equations are real and equal, we need to ensure that the discriminant ([tex]\(\Delta\)[/tex]) of each equation is zero. The discriminant for a general quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
### Equation (i): [tex]\( 4x^2 - 2(k+1)x + (k+4) = 0 \)[/tex]
1. Identify coefficients:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = -2(k+1)\)[/tex]
- [tex]\(c = k + 4\)[/tex]
2. Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the coefficients:
[tex]\[ \Delta = [-2(k+1)]^2 - 4 \cdot 4 \cdot (k+4) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(k+1)^2 - 16(k+4) \][/tex]
[tex]\[ \Delta = 4(k^2 + 2k + 1) - 16k - 64 \][/tex]
[tex]\[ \Delta = 4k^2 + 8k + 4 - 16k - 64 \][/tex]
[tex]\[ \Delta = 4k^2 - 8k - 60 \][/tex]
3. Setting the discriminant to zero:
[tex]\[ 4k^2 - 8k - 60 = 0 \][/tex]
4. Solve the quadratic equation:
Divide the equation by 4:
[tex]\[ k^2 - 2k - 15 = 0 \][/tex]
Factor the quadratic:
[tex]\[ (k-5)(k+3) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ k = 5 \quad \text{or} \quad k = -3 \][/tex]
### Equation (ii): [tex]\( 4x^2 - 2(k+1)x + (k+1) = 0 \)[/tex]
1. Identify coefficients:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = -2(k+1)\)[/tex]
- [tex]\(c = k + 1\)[/tex]
2. Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the coefficients:
[tex]\[ \Delta = [-2(k+1)]^2 - 4 \cdot 4 \cdot (k+1) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(k+1)^2 - 16(k+1) \][/tex]
[tex]\[ \Delta = 4(k^2 + 2k + 1) - 16k - 16 \][/tex]
[tex]\[ \Delta = 4k^2 + 8k + 4 - 16k - 16 \][/tex]
[tex]\[ \Delta = 4k^2 - 8k - 12 \][/tex]
3. Setting the discriminant to zero:
[tex]\[ 4k^2 - 8k - 12 = 0 \][/tex]
4. Solve the quadratic equation:
Divide the equation by 4:
[tex]\[ k^2 - 2k - 3 = 0 \][/tex]
Factor the quadratic:
[tex]\[ (k-3)(k+1) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ k = 3 \quad \text{or} \quad k = -1 \][/tex]
### Summary:
- For equation (i): The values of [tex]\(k\)[/tex] are [tex]\( k = 5 \)[/tex] and [tex]\( k = -3 \)[/tex].
- For equation (ii): The values of [tex]\(k\)[/tex] are [tex]\( k = 3 \)[/tex] and [tex]\( k = -1 \)[/tex].
### Equation (i): [tex]\( 4x^2 - 2(k+1)x + (k+4) = 0 \)[/tex]
1. Identify coefficients:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = -2(k+1)\)[/tex]
- [tex]\(c = k + 4\)[/tex]
2. Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the coefficients:
[tex]\[ \Delta = [-2(k+1)]^2 - 4 \cdot 4 \cdot (k+4) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(k+1)^2 - 16(k+4) \][/tex]
[tex]\[ \Delta = 4(k^2 + 2k + 1) - 16k - 64 \][/tex]
[tex]\[ \Delta = 4k^2 + 8k + 4 - 16k - 64 \][/tex]
[tex]\[ \Delta = 4k^2 - 8k - 60 \][/tex]
3. Setting the discriminant to zero:
[tex]\[ 4k^2 - 8k - 60 = 0 \][/tex]
4. Solve the quadratic equation:
Divide the equation by 4:
[tex]\[ k^2 - 2k - 15 = 0 \][/tex]
Factor the quadratic:
[tex]\[ (k-5)(k+3) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ k = 5 \quad \text{or} \quad k = -3 \][/tex]
### Equation (ii): [tex]\( 4x^2 - 2(k+1)x + (k+1) = 0 \)[/tex]
1. Identify coefficients:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = -2(k+1)\)[/tex]
- [tex]\(c = k + 1\)[/tex]
2. Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the coefficients:
[tex]\[ \Delta = [-2(k+1)]^2 - 4 \cdot 4 \cdot (k+1) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(k+1)^2 - 16(k+1) \][/tex]
[tex]\[ \Delta = 4(k^2 + 2k + 1) - 16k - 16 \][/tex]
[tex]\[ \Delta = 4k^2 + 8k + 4 - 16k - 16 \][/tex]
[tex]\[ \Delta = 4k^2 - 8k - 12 \][/tex]
3. Setting the discriminant to zero:
[tex]\[ 4k^2 - 8k - 12 = 0 \][/tex]
4. Solve the quadratic equation:
Divide the equation by 4:
[tex]\[ k^2 - 2k - 3 = 0 \][/tex]
Factor the quadratic:
[tex]\[ (k-3)(k+1) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ k = 3 \quad \text{or} \quad k = -1 \][/tex]
### Summary:
- For equation (i): The values of [tex]\(k\)[/tex] are [tex]\( k = 5 \)[/tex] and [tex]\( k = -3 \)[/tex].
- For equation (ii): The values of [tex]\(k\)[/tex] are [tex]\( k = 3 \)[/tex] and [tex]\( k = -1 \)[/tex].
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