Answer :
Given the polynomial \(x^2 + 4x - 3\), we know that \(\alpha\) and \(\beta\) are its zeroes.
The sum and product of the zeroes for this polynomial can be given by:
- Sum of zeroes (\(\alpha + \beta\)) = \(-\frac{b}{a} = -\frac{4}{1} = -4\)
- Product of zeroes (\(\alpha \beta\)) = \(\frac{c}{a} = \frac{-3}{1} = -3\)
We need to form a new polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
For the new polynomial:
- The sum of the new zeroes is \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-4}{-3} = \frac{4}{3}\)
- The product of the new zeroes is \(\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-3} = -\frac{1}{3}\)
Now, the quadratic polynomial with zeroes \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is given by:
\[ x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \]
Substituting the sum and product:
\[ x^2 - \left(\frac{4}{3}\right)x - \left(\frac{1}{3}\right) \]
To eliminate the fractions, multiply through by 3:
\[ 3x^2 - 4x - 1 \]
So, the quadratic polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is:
\[ 3x^2 - 4x - 1 \]
The sum and product of the zeroes for this polynomial can be given by:
- Sum of zeroes (\(\alpha + \beta\)) = \(-\frac{b}{a} = -\frac{4}{1} = -4\)
- Product of zeroes (\(\alpha \beta\)) = \(\frac{c}{a} = \frac{-3}{1} = -3\)
We need to form a new polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
For the new polynomial:
- The sum of the new zeroes is \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-4}{-3} = \frac{4}{3}\)
- The product of the new zeroes is \(\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-3} = -\frac{1}{3}\)
Now, the quadratic polynomial with zeroes \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is given by:
\[ x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \]
Substituting the sum and product:
\[ x^2 - \left(\frac{4}{3}\right)x - \left(\frac{1}{3}\right) \]
To eliminate the fractions, multiply through by 3:
\[ 3x^2 - 4x - 1 \]
So, the quadratic polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is:
\[ 3x^2 - 4x - 1 \]
Answer:
Step-by-step explanation:If α and β are the zeros of the polynomial x^2 + 4x - 3, then we can write the polynomial as:
x^2 + 4x - 3 = (x - α)(x - β)
Expanding the right-hand side, we get:
x^2 + 4x - 3 = x^2 - (α + β)x + αβ
Equating the coefficients, we get:
α + β = -4 ... (1)
αβ = -3 ... (2)
Now, we want to form a quadratic polynomial whose zeros are 1/α and 1/β. Let's call this polynomial p(x). Then, we can write:
p(x) = (x - 1/α)(x - 1/β)
Expanding p(x), we get:
p(x) = x^2 - (1/α + 1/β)x + 1/(αβ)
Using equations (1) and (2), we can simplify p(x) as:
p(x) = x^2 + (4/3)x + 1/3
So, the quadratic polynomial whose zeros are 1/α and 1/β is:
p(x) = x^2 + (4/3)x + 1/3
