Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 74615 by chess1 last updated on 27/Nov/19

	Answered by mind is power last updated on 27/Nov/19

	$a = 8 - b$ $(2) \Leftrightarrow {(8 - b)}^{2} + b^{2} + c^{2} = 32$ $x^{2} + y^{2} ⩾ \frac{{(x + y)}^{2}}{2}$ $\Rightarrow {(8 - b)}^{2} + b^{2} ⩾ \frac{{(8 - b + b)}^{2}}{2} = 32$ $but c^{2} + {(8 - b)}^{2} + b^{2} = 32 . . 3$ $\Rightarrow {(8 - b)}^{2} + b^{2} = 32 \Rightarrow 8 - b = b \Rightarrow b = 4$ $3 \Rightarrow c = 0$ $a = 8 - 4 = 4$ $(a, b, c) = (4, 4, 0)$
	Answered by JDamian last updated on 28/Nov/19
	$Geometric approach Expression (2) stands for a sphere which radius length is 4(√(2.)) Expression (1) stands for a plane. The sphere and the plane have contact in c=0. Therefore, the problem is narrowed to find the points where a straight line and a circle intersect: { ((a+b),=,8),((a^2 +b^2 ),=,(32)) :} ⇒ a=4, b=4$
	$Geometric approach$ $E x p r e s s i o n (2) s t a n d s f o r a s p h e r e w h i c h$ $r a d i u s l e n g t h i s 4 \sqrt{2 .}$ $E x p r e s s i o n (1) s t a n d s f o r a p l a n e .$ $T h e s p h e r e a n d t h e p l a n e h a v e c o n t a c t$ $i n c = 0 . T h e r e f o r e, t h e p r o b l e m i s n a r r o w e d$ $t o f i n d t h e p o i n t s w h e r e a s t r a i g h t l i n e a n d$ $a c i r c l e i n t e r s e c t :$ ${\begin{cases} a + b & = & 8 \\ a^{2} + b^{2} & = & 32 \end{cases} \Rightarrow a = 4, b = 4$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum