Understanding the Problem: Finding a New Quadratic Polynomial

The question asks us to find a new quadratic polynomial based on the zeros of a given polynomial. We are given the polynomial \(25x^2 - 15x + 2\), and its zeros are denoted as \(\alpha\) and \(\beta\). We need to find a quadratic polynomial whose zeros are transformed versions of \(\alpha\) and \(\beta\), specifically \((2\alpha)^{-1}\) and \((2\beta)^{-1}\).

Key Concepts for Solving

• For a quadratic polynomial in the form \(ax^2 + bx + c\), the sum of its zeros (\(\alpha + \beta\)) is equal to \(\frac{-b}{a}\).
• The product of its zeros (\(\alpha \beta\)) is equal to \(\frac{c}{a}\).
• A quadratic polynomial with zeros \(r_1\) and \(r_2\) can be written as \(k(x^2 - (r_1 + r_2)x + r_1r_2)\), where \(k\) is any non-zero constant.

Step-by-Step Solution

Step 1: Find the Sum and Product of the Original Zeros (\(\alpha\), \(\beta\))

The given polynomial is \(25x^2 - 15x + 2\). Comparing this to the standard form \(ax^2 + bx + c\), we have:

• \(a = 25\)
• \(b = -15\)
• \(c = 2\)

Using the formulas for the sum and product of zeros:

Sum of zeros: \(\alpha + \beta = \frac{-b}{a} = \frac{-(-15)}{25} = \frac{15}{25} = \frac{3}{5}\)

Product of zeros: \(\alpha \beta = \frac{c}{a} = \frac{2}{25}\)

Step 2: Determine the New Zeros

The new zeros are given as \((2\alpha)^{-1}\) and \((2\beta)^{-1}\). We can rewrite these as \(\frac{1}{2\alpha}\) and \(\frac{1}{2\beta}\).

Step 3: Find the Sum of the New Zeros

Let the new zeros be \(r_1 = \frac{1}{2\alpha}\) and \(r_2 = \frac{1}{2\beta}\). The sum of the new zeros is:

Sum = \(r_1 + r_2 = \frac{1}{2\alpha} + \frac{1}{2\beta}\)

To add these fractions, we find a common denominator, which is \(2\alpha\beta\):

Sum = \(\frac{\beta}{2\alpha\beta} + \frac{\alpha}{2\alpha\beta} = \frac{\alpha + \beta}{2\alpha\beta}\)

Now, substitute the values we found for \(\alpha + \beta\) and \(\alpha\beta\) in Step 1:

Sum = \(\frac{3/5}{2(2/25)} = \frac{3/5}{4/25}\)

To divide fractions, we multiply by the reciprocal of the second fraction:

Sum = \(\frac{3}{5} \times \frac{25}{4} = \frac{3 \times 25}{5 \times 4} = \frac{75}{20} = \frac{15}{4}\) (by dividing numerator and denominator by 5)

Step 4: Find the Product of the New Zeros

The product of the new zeros is:

Product = \(r_1 \times r_2 = \frac{1}{2\alpha} \times \frac{1}{2\beta}\)

Multiply the numerators and the denominators:

Product = \(\frac{1 \times 1}{(2\alpha) \times (2\beta)} = \frac{1}{4\alpha\beta}\)

Substitute the value of \(\alpha\beta\) from Step 1:

Product = \(\frac{1}{4(2/25)} = \frac{1}{8/25}\)

Multiply by the reciprocal:

Product = \(1 \times \frac{25}{8} = \frac{25}{8}\)

Step 5: Construct the New Quadratic Polynomial

A quadratic polynomial with sum of zeros \(S = \frac{15}{4}\) and product of zeros \(P = \frac{25}{8}\) is given by \(k(x^2 - Sx + P)\), where \(k\) is a non-zero constant.

The polynomial is \(k\left(x^2 - \frac{15}{4}x + \frac{25}{8}\right)\).

To find a polynomial with integer coefficients, we can choose \(k\) to be the least common multiple (LCM) of the denominators 4 and 8, which is 8. Let \(k=8\).

New polynomial = \(8\left(x^2 - \frac{15}{4}x + \frac{25}{8}\right)\)

Distribute the 8:

New polynomial = \(8 \times x^2 - 8 \times \frac{15}{4}x + 8 \times \frac{25}{8}\)

New polynomial = \(8x^2 - \left(\frac{8 \times 15}{4}\right)x + \left(\frac{8 \times 25}{8}\right)\)

New polynomial = \(8x^2 - (2 \times 15)x + 25\)

New polynomial = \(8x^2 - 30x + 25\)

Conclusion

The quadratic polynomial whose zeros are \((2\alpha)^{-1}\) and \((2\beta)^{-1}\) is \(8x^2 - 30x + 25\). This matches one of the given options.

Revision Table: Quadratic Polynomial Zeros

Here's a summary of the key formulas used:

• For \(ax^2 + bx + c\), sum of zeros \(= -\frac{b}{a}\)
• For \(ax^2 + bx + c\), product of zeros \(= \frac{c}{a}\)
• Polynomial with sum of zeros \(S\) and product of zeros \(P\) is \(k(x^2 - Sx + P)\)

Additional Information: Transformations of Zeros

Problems involving transformations of zeros are common in quadratic equations. If a polynomial has zeros \(\alpha\) and \(\beta\), and we want a new polynomial with zeros \(f(\alpha)\) and \(f(\beta)\), the general approach is to:

• Find the sum and product of the original zeros using the coefficients of the given polynomial.
• Calculate the sum and product of the new zeros in terms of \(\alpha\) and \(\beta\).
• Substitute the known values of \(\alpha + \beta\) and \(\alpha \beta\) into the expressions for the sum and product of the new zeros.
• Use the formula \(k(x^2 - (\text{sum of new zeros})x + (\text{product of new zeros}))\) to construct the new polynomial.

In this specific problem, the transformation is \(f(z) = (2z)^{-1} = \frac{1}{2z}\). Other common transformations might involve \(z+k\), \(kz\), \(z^2\), \(1/z\), etc. The method remains similar: find the new sum and product in terms of the old sum and product.