Multiple choice

If $\alpha $ and $\beta $ are the two zeros of the polynomial ${ 25p }^{ 2 }-15p+2,$ find a quadratic polynomial whose zeros are $\dfrac { 1 }{ 2\alpha }$ and $\dfrac { 1 }{ 2\beta } .$

  1. $\dfrac { 1 }{ 8 } \left( { 6p }^{ 2 }+25p-30 \right) $
  2. $\dfrac { 1 }{ 8 } \left( { 6p }^{ 2 }-30p+25 \right) $
  3. $\dfrac { 1 }{ 8 } \left( { 8p }^{ 2 }+30p-25 \right) $
  4. $\dfrac { 1 }{ 8 } \left( { 8p }^{ 2 }-30p+25 \right) $
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D Correct answer
Explanation

For 25p^2 - 15p + 2 = 0, the sum of zeros alpha + beta = 15/25 = 3/5 and product alpha*beta = 2/25. The new zeros are 1/(2alpha) and 1/(2beta). Their sum is (alpha+beta)/(2*alpha*beta) = (3/5) / (4/25) = 15/4, and their product is 1/(4*alpha*beta) = 1/(8/25) = 25/8. The polynomial is x^2 - (15/4)x + 25/8, which scales to (1/8)(8x^2 - 30x + 25).

AI explanation

Using the relationships between roots and coefficients for the polynomial 25p^2 - 15p + 2, the sum of the roots alpha + beta is 15/25 = 3/5 and the product alpha * beta is 2/25. The sum of the new zeros 1/(2*alpha) + 1/(2*beta) equals (alpha + beta) / (2*alpha*beta) = (3/5) / (4/25) = 15/4. The product of the new zeros is 1 / (4*alpha*beta) = 25/8. A valid quadratic polynomial with these zeros is p^2 - (15/4)p + 25/8, which is equivalent to (1/8)*(8p^2 - 30p + 25).

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