Given a,b,c and d are reals numbers such that { ((a^2 +b^2 =10)),((c^2 +d^2 =10 )),((ab+cd=0)) :} Find ac + bd.


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Question Number 212499 by efronzo1 last updated on 15/Oct/24
$Given a,b,c and d are reals numbers such that { ((a^2 +b^2 =10)),((c^2 +d^2 =10 )),((ab+cd=0)) :} Find ac + bd.$
$Given a, b, c and d are reals$ $numbers such that$ ${\begin{cases} a^{2} + b^{2} = 10 \\ c^{2} + d^{2} = 10 \\ ab + cd = 0 \end{cases}$ $Find ac + bd .$
	Answered by ajfour last updated on 15/Oct/24

	${(a + b)}^{2} = 10 + 2 a b$ ${(c + d)}^{2} = 10 + 2 c d$ $l e t \frac{d}{b} = - \frac{c}{a} = k$ $c^{2} + d^{2} = k^{2} (a^{2} + b^{2}) = 10$ $\Rightarrow k^{2} = 1 a s (a^{2} + b^{2}) = 10$ $s a y k = 1 \Rightarrow$ $d = b, c = - a$ $a c + b d = - a^{2} + b^{2} = 10 - 2 a^{2} = 2 b^{2} - 10$ $i t h i n k i t d e p e n d s o n c h o i c e o f$ $a^{2}, b^{2} w i t h t h e c o n d i t i o n a^{2} + b^{2} = 10$
	Answered by Ghisom last updated on 16/Oct/24

	$a b + c d = 0 \Rightarrow d = - \frac{a b}{c}$ $c^{2} + d^{2} = 10 \Rightarrow b^{2} = \frac{c^{2} (10 - c^{2})}{a^{2}}$ $a^{2} + b^{2} = 10 \Rightarrow a^{4} - 10 a^{2} - c^{4} + 10 c^{2} = 0$ $\Rightarrow$ $c = \pm a \lor c = \pm \sqrt{10 - a^{2}}$ $\Rightarrow a c + b d = \pm 2 (a^{2} - 5) \lor a c + b d = 0$
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