Question

If p & q are the roots of the Equation x2-bx + C = 0, then what is the Equation whose roots are (pq + p + q) and (pq - p-a)? (a) x2 - 2cx + C2 - b2 = 0 (b) x2 - 2bx + C2 + b2 = 0 (c) 8cx- 2(b + c)X + C2 = 0 (d) x2 + 2bx - (C2-b2) = 0​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$so \: given \: quadratic \: equation \: is \\ x {}^{2} - bx + c = 0 \\ whose \: roots \: are \: p \: and \: q \\ \\ let \: then \\ \alpha = p \: \\ \beta = q \\ \\ comparing \: it \: with \\ ax {}^{2} + bx + c = 0 \\ we \: have \\ \\ a = 1 \\ b = - b\\ c = c \\ \\ we \: know \: that \\ \\ \alpha + \beta = \frac{ - b}{a} \\ \\ p + q = \frac{ - ( - b)}{1} \\ \\ p +q = b \: \: \: \: \: \: \: \: (1) \\ \\ similarly \\ \\ \alpha \beta = \frac{c}{a} \\ \\ pq = \frac{c}{1} \\ \\ pq = c \: \: \: \: \: \: \: \: \: \: \: \: (2)$

$so \: thus \: then \: \\ for \: a \: certain \: quadratic \: equation \\ its \: two \: \: respective \: roots \: are \\ \\ ( \: pq + p + q) \: and \: \: (pq - p - q) \\ \\ thus \: then \\ \\ from \: \: (1) \: \: and \: \: (2) \\ we \: get \\ \\( \: pq + p + q \: ) = b + c \\ \\ pq - p - q = pq - (p + q) \\ \\ = c - b$

$thus \: then \: let \: here \\ \\ b + c = \alpha \\ \\ c - b = \beta \\ \\ we \: know \: that \\ formation \: of \: \: any \: quadratic \: equation \: \\ is \: given \: \: by \\ \\ x {}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ \\ x {}^{2} - (b + c + c - b)x + (b + c)(c - b) = 0 \\ \\ = x {}^{2} - (2c)x \: + ( - b { }^{2} + c {}^{2} ) \\ \\ x {}^{2} - 2cx + c {}^{2} - b { }^{2} = 0 \\ \\ therefore \: \: our \: required \\ quadratic \: \: equation \: \: is \\ \\ x {}^{2} - 2cx + c {}^{2} - b {}^{2} = 0$