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Sagot :
Given the quadratic polynomial [tex]\( x^2 - x - 30 \)[/tex], we need to find its zeroes and verify the relationship between the zeroes and its coefficients.
### Step-by-Step Solution
1. Identify the coefficients:
The given quadratic polynomial is [tex]\( x^2 - x - 30 \)[/tex]. Here:
[tex]\[ a = 1, \quad b = -1, \quad c = -30 \][/tex]
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-1)^2 - 4(1)(-30) = 1 + 120 = 121 \][/tex]
3. Find the roots using the quadratic formula:
The roots [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex] of the quadratic equation are given by:
[tex]\[ x_{1,2} = \frac{{-b \pm \sqrt{\Delta}}}{2a} \][/tex]
Substituting [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{{-(-1) + \sqrt{121}}}{2 \times 1} = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ x_2 = \frac{{-(-1) - \sqrt{121}}}{2 \times 1} = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
So, the zeroes of the polynomial are [tex]\( x_1 = 6 \)[/tex] and [tex]\( x_2 = -5 \)[/tex].
4. Verify the relationship between the zeroes and the coefficients:
According to the quadratic polynomial [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum and product of the zeroes should satisfy the following relationships:
- Sum of the zeroes: [tex]\( x_1 + x_2 = -\frac{b}{a} \)[/tex]
- Product of the zeroes: [tex]\( x_1 \cdot x_2 = \frac{c}{a} \)[/tex]
Sum of the zeroes:
[tex]\[ x_1 + x_2 = 6 + (-5) = 1 \][/tex]
[tex]\[ -\frac{b}{a} = -\frac{-1}{1} = 1 \][/tex]
Hence, the sum of the zeroes [tex]\(1\)[/tex] is equal to [tex]\(-\frac{b}{a}\)[/tex], which is also [tex]\(1\)[/tex].
Product of the zeroes:
[tex]\[ x_1 \cdot x_2 = 6 \cdot (-5) = -30 \][/tex]
[tex]\[ \frac{c}{a} = \frac{-30}{1} = -30 \][/tex]
Hence, the product of the zeroes [tex]\(-30\)[/tex] is equal to [tex]\(\frac{c}{a}\)[/tex], which is also [tex]\(-30\)[/tex].
### Conclusion
The zeroes of the quadratic polynomial [tex]\( x^2 - x - 30 \)[/tex] are [tex]\(6\)[/tex] and [tex]\(-5\)[/tex]. The sum of the zeroes [tex]\((6 + (-5) = 1)\)[/tex] confirms the relationship [tex]\(-\frac{b}{a} = 1\)[/tex], and the product of the zeroes [tex]\((6 \cdot (-5) = -30)\)[/tex] confirms the relationship [tex]\(\frac{c}{a} = -30\)[/tex].
### Step-by-Step Solution
1. Identify the coefficients:
The given quadratic polynomial is [tex]\( x^2 - x - 30 \)[/tex]. Here:
[tex]\[ a = 1, \quad b = -1, \quad c = -30 \][/tex]
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-1)^2 - 4(1)(-30) = 1 + 120 = 121 \][/tex]
3. Find the roots using the quadratic formula:
The roots [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex] of the quadratic equation are given by:
[tex]\[ x_{1,2} = \frac{{-b \pm \sqrt{\Delta}}}{2a} \][/tex]
Substituting [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{{-(-1) + \sqrt{121}}}{2 \times 1} = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ x_2 = \frac{{-(-1) - \sqrt{121}}}{2 \times 1} = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
So, the zeroes of the polynomial are [tex]\( x_1 = 6 \)[/tex] and [tex]\( x_2 = -5 \)[/tex].
4. Verify the relationship between the zeroes and the coefficients:
According to the quadratic polynomial [tex]\( ax^2 + bx + c = 0 \)[/tex], the sum and product of the zeroes should satisfy the following relationships:
- Sum of the zeroes: [tex]\( x_1 + x_2 = -\frac{b}{a} \)[/tex]
- Product of the zeroes: [tex]\( x_1 \cdot x_2 = \frac{c}{a} \)[/tex]
Sum of the zeroes:
[tex]\[ x_1 + x_2 = 6 + (-5) = 1 \][/tex]
[tex]\[ -\frac{b}{a} = -\frac{-1}{1} = 1 \][/tex]
Hence, the sum of the zeroes [tex]\(1\)[/tex] is equal to [tex]\(-\frac{b}{a}\)[/tex], which is also [tex]\(1\)[/tex].
Product of the zeroes:
[tex]\[ x_1 \cdot x_2 = 6 \cdot (-5) = -30 \][/tex]
[tex]\[ \frac{c}{a} = \frac{-30}{1} = -30 \][/tex]
Hence, the product of the zeroes [tex]\(-30\)[/tex] is equal to [tex]\(\frac{c}{a}\)[/tex], which is also [tex]\(-30\)[/tex].
### Conclusion
The zeroes of the quadratic polynomial [tex]\( x^2 - x - 30 \)[/tex] are [tex]\(6\)[/tex] and [tex]\(-5\)[/tex]. The sum of the zeroes [tex]\((6 + (-5) = 1)\)[/tex] confirms the relationship [tex]\(-\frac{b}{a} = 1\)[/tex], and the product of the zeroes [tex]\((6 \cdot (-5) = -30)\)[/tex] confirms the relationship [tex]\(\frac{c}{a} = -30\)[/tex].
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