If α and β are the zeros of the quadratic polynomial f(x) = x² − 3x − 2, find the quadratic polynomial whose zeros are 1/(2α + β) and 1/(2β + α)

Video Explanation

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Solution

Given polynomial:

f(x) = x² − 3x − 2

Step 1: Find α + β and αβ

Comparing f(x) = x² − 3x − 2 with ax² + bx + c:

a = 1,   b = −3,   c = −2

α + β = −b/a = 3

αβ = c/a = −2

Step 2: Find the Sum of the New Zeros

Sum of new zeros:

1/(2α + β) + 1/(2β + α)

= (2β + α + 2α + β) / (2α + β)(2β + α)

= 3(α + β) / (2α + β)(2β + α)

Now,

(2α + β)(2β + α)

= 5αβ + 2(α² + β²)

α² + β² = (α + β)² − 2αβ

= 9 − 2(−2)

= 13

∴ (2α + β)(2β + α) = 5(−2) + 2(13)

= −10 + 26

= 16

∴ Sum of new zeros = 3 × 3 / 16 = 9/16

Step 3: Find the Product of the New Zeros

Product of new zeros:

1 / (2α + β)(2β + α)

= 1/16

Step 4: Form the Required Quadratic Polynomial

The quadratic polynomial whose zeros are 1/(2α + β) and 1/(2β + α) is:

x² − (sum of zeros)x + (product of zeros)

= x² − (9/16)x + 1/16

Multiplying throughout by 16:

16x² − 9x + 1

Final Answer

The required quadratic polynomial is 16x² − 9x + 1.

Conclusion

Thus, the quadratic polynomial whose zeros are 1/(2α + β) and 1/(2β + α) is 16x² − 9x + 1.