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Sagot :
Let's solve each part of the question step-by-step.
### Problem 17: Find [tex]\( k \)[/tex] so that [tex]\( (k-12)x^2 + 2(k-12)x + 2 = 0 \)[/tex] has equal roots. [tex]\( (k \neq 12) \)[/tex]
For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] to have equal roots, its discriminant must be zero. The discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here, the equation is [tex]\( (k-12)x^2 + 2(k-12)x + 2 = 0 \)[/tex].
- [tex]\( a = k-12 \)[/tex]
- [tex]\( b = 2(k-12) \)[/tex]
- [tex]\( c = 2 \)[/tex]
We set up the discriminant equation:
[tex]\[ \Delta = \left[2(k-12)\right]^2 - 4(k-12) \cdot 2 = 0 \][/tex]
Expanding this equation:
[tex]\[ \left[2(k-12)\right]^2 - 4 \cdot 2 \cdot (k-12) = 0 \][/tex]
[tex]\[ 4(k-12)^2 - 8(k-12) = 0 \][/tex]
Factor out [tex]\( 4(k-12) \)[/tex] from the equation:
[tex]\[ 4(k-12)\left[(k-12) - 2\right] = 0 \][/tex]
This gives us two possible solutions:
[tex]\[ 4(k-12) = 0 \quad \text{or} \quad (k-12) - 2 = 0 \][/tex]
Since [tex]\( k \neq 12 \)[/tex], the only valid solution is:
[tex]\[ (k - 12) - 2 = 0 \][/tex]
[tex]\[ k - 14 = 0 \][/tex]
Thus,
[tex]\[ k = 14 \][/tex]
### Problem 18: Find the nature of the roots of the following quadratic equations:
#### (a) [tex]\( 5x^2 - 3x + 2 = 0 \)[/tex]
For this quadratic equation, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here:
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = -3 \)[/tex]
- [tex]\( c = 2 \)[/tex]
Substitute the values into the discriminant formula:
[tex]\[ \Delta = (-3)^2 - 4 \cdot 5 \cdot 2 \][/tex]
[tex]\[ \Delta = 9 - 40 \][/tex]
[tex]\[ \Delta = -31 \][/tex]
Since [tex]\(\Delta < 0\)[/tex], the roots are complex.
#### (b) [tex]\( 2x^2 - x - 6 = 0 \)[/tex]
Here:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -1 \)[/tex]
- [tex]\( c = -6 \)[/tex]
Calculate the discriminant:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 2 \cdot (-6) \][/tex]
[tex]\[ \Delta = 1 + 48 \][/tex]
[tex]\[ \Delta = 49 \][/tex]
Since [tex]\(\Delta > 0\)[/tex], the roots are real and unequal.
#### (c) [tex]\( x^2 + 6x + 9 = 0 \)[/tex]
Here:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 6 \)[/tex]
- [tex]\( c = 9 \)[/tex]
Calculate the discriminant:
[tex]\[ \Delta = 6^2 - 4 \cdot 1 \cdot 9 \][/tex]
[tex]\[ \Delta = 36 - 36 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
Since [tex]\(\Delta = 0\)[/tex], the roots are real and equal.
### Summary:
- For problem 17, the value of [tex]\( k \)[/tex] is [tex]\( 14 \)[/tex].
- For problem 18:
- (a) The roots are complex.
- (b) The roots are real and unequal.
- (c) The roots are real and equal.
### Problem 17: Find [tex]\( k \)[/tex] so that [tex]\( (k-12)x^2 + 2(k-12)x + 2 = 0 \)[/tex] has equal roots. [tex]\( (k \neq 12) \)[/tex]
For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] to have equal roots, its discriminant must be zero. The discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here, the equation is [tex]\( (k-12)x^2 + 2(k-12)x + 2 = 0 \)[/tex].
- [tex]\( a = k-12 \)[/tex]
- [tex]\( b = 2(k-12) \)[/tex]
- [tex]\( c = 2 \)[/tex]
We set up the discriminant equation:
[tex]\[ \Delta = \left[2(k-12)\right]^2 - 4(k-12) \cdot 2 = 0 \][/tex]
Expanding this equation:
[tex]\[ \left[2(k-12)\right]^2 - 4 \cdot 2 \cdot (k-12) = 0 \][/tex]
[tex]\[ 4(k-12)^2 - 8(k-12) = 0 \][/tex]
Factor out [tex]\( 4(k-12) \)[/tex] from the equation:
[tex]\[ 4(k-12)\left[(k-12) - 2\right] = 0 \][/tex]
This gives us two possible solutions:
[tex]\[ 4(k-12) = 0 \quad \text{or} \quad (k-12) - 2 = 0 \][/tex]
Since [tex]\( k \neq 12 \)[/tex], the only valid solution is:
[tex]\[ (k - 12) - 2 = 0 \][/tex]
[tex]\[ k - 14 = 0 \][/tex]
Thus,
[tex]\[ k = 14 \][/tex]
### Problem 18: Find the nature of the roots of the following quadratic equations:
#### (a) [tex]\( 5x^2 - 3x + 2 = 0 \)[/tex]
For this quadratic equation, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here:
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = -3 \)[/tex]
- [tex]\( c = 2 \)[/tex]
Substitute the values into the discriminant formula:
[tex]\[ \Delta = (-3)^2 - 4 \cdot 5 \cdot 2 \][/tex]
[tex]\[ \Delta = 9 - 40 \][/tex]
[tex]\[ \Delta = -31 \][/tex]
Since [tex]\(\Delta < 0\)[/tex], the roots are complex.
#### (b) [tex]\( 2x^2 - x - 6 = 0 \)[/tex]
Here:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -1 \)[/tex]
- [tex]\( c = -6 \)[/tex]
Calculate the discriminant:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 2 \cdot (-6) \][/tex]
[tex]\[ \Delta = 1 + 48 \][/tex]
[tex]\[ \Delta = 49 \][/tex]
Since [tex]\(\Delta > 0\)[/tex], the roots are real and unequal.
#### (c) [tex]\( x^2 + 6x + 9 = 0 \)[/tex]
Here:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 6 \)[/tex]
- [tex]\( c = 9 \)[/tex]
Calculate the discriminant:
[tex]\[ \Delta = 6^2 - 4 \cdot 1 \cdot 9 \][/tex]
[tex]\[ \Delta = 36 - 36 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
Since [tex]\(\Delta = 0\)[/tex], the roots are real and equal.
### Summary:
- For problem 17, the value of [tex]\( k \)[/tex] is [tex]\( 14 \)[/tex].
- For problem 18:
- (a) The roots are complex.
- (b) The roots are real and unequal.
- (c) The roots are real and equal.
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