Question

solve for x: x2-2(a+2)x+(a+1)(a+2)=0

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ 2 \: possible \: values \: of \: x \\ \\ so \: here \\ \: given \: quadratic \: equation \: is \\ x {}^{2} - 2(a + 2)x + (a + 1)(a + 2) = 0 \\ \\ comparing \: it \: with \\ \: ax {}^{2} + bx + c = 0 \\ we \: have \\ \\ a = 1 \\ b = - 2(a + 2) \\ c = (a + 1)(a + 2) \\ \\ so \: using \\ Δ = b {}^{2} - 4ac \\ \\ = ( - 2(a + 2)) {}^{2} - 4 \times 1 \times (a + 1)(a + 2) \\ \\ = 4(a + 2) {}^{2} - 4(a {}^{2} + 2a + a + 2) \\ \\ = 4(a {}^{2} + 4 + 4a) - 4(a {}^{2} + 3a + 2) \\ \\ = 4(a {}^{2} + 4 + 4a - (a {}^{2} + 3a + 2) \\ \\ = 4(a {}^{2} + 4 + 4a - a {}^{2} - 3a - 2) \\ \\ Δ = 4(a + 2)$

$so \: by \: formula \: method \\ we \: get \\ \\ x = \frac{ - b \:± \: \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - ( - 2(a + 2)) \: ± \: \sqrt{4(a + 2)} }{(2 \times 1)} \\ \\ = \frac{2(a + 2) \:± \: 2 \sqrt{(a + 2)} }{2} \\ \\ x= a + 2 \: ± \: \sqrt{(a + 2)}$