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If alpha and beta are zeroes of the polynomial x square 4 x - 3 then form a quadratic polynomial whose zeroes are 1by alpha and 1 by beta



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:
\[ 3x^2 - 4x - 1 \]

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