are the roots of the equation \( \mathrm{x}^{2}-\mathrm{px}+\mathrm{q}=0 \), then the equation whose roots are \( \frac{q}{p-\alpha} \) and \( \frac{q}{p-\beta} \)
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To find the equation whose roots are \( \frac{q}{p-\alpha} \) and \( \frac{q}{p-\beta} \), where \(\alpha\) and \(\beta\) are the roots of the original equation \( x^{2}-px+q=0 \), we can first use Vieta's formulas. According to Vieta's, \( \alpha + \beta = p \) and \( \alpha \beta = q \). Now, we can calculate the sum and product of the new roots: 1. The sum of the new roots: \[ \frac{q}{p-\alpha} + \frac{q}{p-\beta} = q \left( \frac{1}{p-\alpha} + \frac{1}{p-\beta} \right) = q \left( \frac{(p-\beta) + (p-\alpha)}{(p-\alpha)(p-\beta)} \right) = q \left( \frac{2p - (\alpha + \beta)}{(p-\alpha)(p-\beta)} \right) \] Substitute \(\alpha + \beta = p\): \[ = q \left( \frac{2p - p}{(p-\alpha)(p-\beta)} \right) = q \left( \frac{p}{(p-\alpha)(p-\beta)} \right) \] Now, \( (p-\alpha)(p-\beta) = p^2 - p(\alpha + \beta) + \alpha \beta = p^2 - p^2 + q = q \). Therefore, the sum is: \[ \frac{qp}{q} = p \] 2. The product of the new roots: \[ \frac{q}{p-\alpha} \cdot \frac{q}{p-\beta} = \frac{q^2}{(p-\alpha)(p-\beta)} = \frac{q^2}{q} = q \] Thus, the new equation with roots \( \frac{q}{p-\alpha} \) and \( \frac{q}{p-\beta} \) will have the form \( x^2 - (p)x + q = 0 \) which simplifies to: \[ x^2 - px + q = 0 \] This means that the structure of the new equation reflects the original equation!