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Question-27050

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Question Number 27050 by Joel578 last updated on 01/Jan/18
Commented by Joel578 last updated on 01/Jan/18
(a + c)^2 − (b + d)^2 = a^2 + c^2 + 2ac − b^2 − d^2 − 2bd = (a^2 + c^2 ) − (b^2 + d^2 ) + 2(ac − bd) 2(ac − bd) = (a + c)^2 − (b + d)^2 + (b^2 + d^2 ) − (a^2 + c^2 ) (b − a)(b + a) + (d − c)(d + c) = 2(ac − bd) (b^2 − a^2 ) + (d^2 − c^2 ) = (a + c)^2 − (b + d)^2 + (b^2 + d^2 ) − (a^2 + c^2 ) (b^2 + d^2 ) − (a^2 + c^2 ) = (a + c)^2 − (b + d)^2 + (b^2 + d^2 ) − (a^2 + c^2 ) (a + c)^2 − (b + d)^2 = 0 (a + c + b + d)(a + c − b − d) = 0 a + c + b + d = 0 ∨ a + c − b − d = 0 (a + b + c + d)^2 = 0
$$\left({a}\:+\:{c}\right)^{\mathrm{2}} \:−\:\left({b}\:+\:{d}\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{ac}\:−\:{b}^{\mathrm{2}} \:−\:{d}^{\mathrm{2}} \:−\:\mathrm{2}{bd} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right)\:−\:\left({b}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \right)\:+\:\mathrm{2}\left({ac}\:−\:{bd}\right) \\ $$$$\mathrm{2}\left({ac}\:−\:{bd}\right)\:=\:\left({a}\:+\:{c}\right)^{\mathrm{2}} \:−\:\left({b}\:+\:{d}\right)^{\mathrm{2}} \:+\:\left({b}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\left({b}\:−\:{a}\right)\left({b}\:+\:{a}\right)\:+\:\left({d}\:−\:{c}\right)\left({d}\:+\:{c}\right)\:=\:\mathrm{2}\left({ac}\:−\:{bd}\right) \\ $$$$\left({b}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \right)\:+\:\left({d}^{\mathrm{2}} \:−\:{c}^{\mathrm{2}} \right)\:=\:\left({a}\:+\:{c}\right)^{\mathrm{2}} \:−\:\left({b}\:+\:{d}\right)^{\mathrm{2}} \:+\:\left({b}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right) \\ $$$$\left({b}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right)\:=\:\left({a}\:+\:{c}\right)^{\mathrm{2}} \:−\:\left({b}\:+\:{d}\right)^{\mathrm{2}} \:+\:\left({b}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right) \\ $$$$\left({a}\:+\:{c}\right)^{\mathrm{2}} \:−\:\left({b}\:+\:{d}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left({a}\:+\:{c}\:+\:{b}\:+\:{d}\right)\left({a}\:+\:{c}\:−\:{b}\:−\:{d}\right)\:=\:\mathrm{0} \\ $$$${a}\:+\:{c}\:+\:{b}\:+\:{d}\:=\:\mathrm{0}\:\:\:\vee\:\:{a}\:+\:{c}\:−\:{b}\:−\:{d}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\left({a}\:+\:{b}\:+\:{c}\:+\:{d}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$
Commented by Joel578 last updated on 01/Jan/18
Is there any way to solve using the given function f(x) ?
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{function}\:{f}\left({x}\right)\:? \\ $$

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