Answer :
Sure! Let's delve into solving what the sum and product of the zeroes of a quadratic polynomial are.
Given a quadratic polynomial in the form:
[tex]\[ ax^2 + bx + c \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are constants with [tex]\(a \neq 0\)[/tex], the zeroes (also known as roots) of the polynomial are typically denoted by [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
### Step 1: Sum of the Zeroes
The sum of the zeroes of the quadratic polynomial can be determined using the relationship from Vieta's formulas, which tells us:
[tex]\[ \alpha + \beta = -\frac{b}{a} \][/tex]
### Step 2: Product of the Zeroes
The product of the zeroes of the quadratic polynomial can also be determined using Vieta's formulas, which gives:
[tex]\[ \alpha \beta = \frac{c}{a} \][/tex]
### Example with specific values
Let's consider specific values to illustrate this concept:
- Let [tex]\( a = 2 \)[/tex]
- Let [tex]\( b = 3 \)[/tex]
- Let [tex]\( c = 1 \)[/tex]
Using these values:
1. Sum of the Zeroes:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{3}{2} = -1.5 \][/tex]
2. Product of the Zeroes:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{1}{2} = 0.5 \][/tex]
So, for the quadratic polynomial with the coefficients [tex]\(a = 2\)[/tex], [tex]\( b = 3\)[/tex], and [tex]\(c = 1\)[/tex]:
- The sum of the zeroes is [tex]\(-1.5\)[/tex]
- The product of the zeroes is [tex]\(0.5\)[/tex]
Therefore, the sum and product of the zeroes [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] of the quadratic polynomial [tex]\(2x^2 + 3x + 1\)[/tex] are:
[tex]\[ \alpha + \beta = -1.5 , \quad \alpha \beta = 0.5 \][/tex]
Given a quadratic polynomial in the form:
[tex]\[ ax^2 + bx + c \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are constants with [tex]\(a \neq 0\)[/tex], the zeroes (also known as roots) of the polynomial are typically denoted by [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
### Step 1: Sum of the Zeroes
The sum of the zeroes of the quadratic polynomial can be determined using the relationship from Vieta's formulas, which tells us:
[tex]\[ \alpha + \beta = -\frac{b}{a} \][/tex]
### Step 2: Product of the Zeroes
The product of the zeroes of the quadratic polynomial can also be determined using Vieta's formulas, which gives:
[tex]\[ \alpha \beta = \frac{c}{a} \][/tex]
### Example with specific values
Let's consider specific values to illustrate this concept:
- Let [tex]\( a = 2 \)[/tex]
- Let [tex]\( b = 3 \)[/tex]
- Let [tex]\( c = 1 \)[/tex]
Using these values:
1. Sum of the Zeroes:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{3}{2} = -1.5 \][/tex]
2. Product of the Zeroes:
[tex]\[ \alpha \beta = \frac{c}{a} = \frac{1}{2} = 0.5 \][/tex]
So, for the quadratic polynomial with the coefficients [tex]\(a = 2\)[/tex], [tex]\( b = 3\)[/tex], and [tex]\(c = 1\)[/tex]:
- The sum of the zeroes is [tex]\(-1.5\)[/tex]
- The product of the zeroes is [tex]\(0.5\)[/tex]
Therefore, the sum and product of the zeroes [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] of the quadratic polynomial [tex]\(2x^2 + 3x + 1\)[/tex] are:
[tex]\[ \alpha + \beta = -1.5 , \quad \alpha \beta = 0.5 \][/tex]
