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SMC2012_web_solutions

So here 1 a 4 y b and 16 4 2 y y c we see that 16 4 4

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So here 1 = a , 4 + = y b and 16 4 2 + = y y c . We see that ) 16 4 ( 4 ) 4 ( 4 2 2 2 + + = y y y ac b 2 2 2 ) 4 ( 3 ) 16 8 ( 3 48 24 3 = + = + = y y y y y . So there is a real number solution for x if and only if 0 ) 4 ( 3 2 y . Now, as for all real numbers y , 0 ) 4 ( 2 y , it follows that 4 0 4 0 ) 4 ( 3 2 = = y y y . When 4 = y the quadratic equation becomes 0 16 8 2 = + + x x , that is 0 ) 4 ( 2 = + x , which has just the one solution 4 = x . So ) 4 , 4 ( is the only pair of real numbers, ) , ( y x , for which ) 4 )( 4 ( ) ( 2 + = + y x y x . Method 2. We can simplify the algebra by making the substitution 4 + = x w and 4 = y z . Then z w z w y x + = + + = + ) 4 ( ) 4 ( , and the equation becomes wz z w = + 2 ) ( . Now, 0 ) ( ) ( 0 2 ) ( 2 2 3 2 2 1 2 2 2 2 2 = + + = + + = + + = + z z w z wz w wz z wz w wz z w . The sum of the squares of two real numbers is zero if and only if each real number is 0. So the only real number solution of 0 ) ( ) ( 2 2 3 2 2 1 = + + z z w is 0 2 3 2 1 = = + z z w . This is equivalent to 0 = = z w and hence to 4 = x and 4 = y . So again we deduce that there is just this one solution. Extension Problems 25.1 It is possible to use the “ ac b 4 2 ” criterion to show that ) 0 , 0 ( is the only pair of real numbers that satisfy the equation 0 2 2 = + + z wz w . Check this. 25.2 Show that the “ ac b 4 2 ” criterion is correct by proving that for all real numbers, a , b , c , with 0 a , the quadratic equation 0 2 = + + c bx ax has a real number solution if and only if ac b 4 2 . 25.3 [For those who know about complex numbers.] If we allow the possibility that x and y are complex numbers, then the equation ) 4 )( 4 ( ) ( 2 + = + y x y x has more than one solution. Check that 6 = x , i y 3 5 1 + = is one solution of this equation. How many more can you find?
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2 0 . E T h e t w o t a n g e n t s d r a w n f r o m a p o i n t o u t s i d e a c i r c l e t o t h a t c i r c l e a r e e q u a l i n l e n g t h . T h i s t h e o r e m h a s b e e n u s e d t o m a r k f o u r p a i r s o f e q u a l l i n e s e g m e n t s o n t h e d i a g r a m . I n t h e c i r c l e t h e d i a m e t e r , , h a s b e e n m a r k e d . I t i s a l s o a p e r p e n d i c u l a r h e i g h t o f t h e t r a p e z i u m . X Y W e a r e g i v e n t h a t s o w e c a n d e d u c e t h a t . T h e a r e a o f t r a p e z i u m . T h e r e f o r e S R = P Q = 2 5 c m ( a + d ) + ( b + c ) = 2 5 + 2 5 = 5 0 P Q R S = 1 2 ( S P + Q R ) × X Y = 6 0 0 c m 2 P Q R S C X Y a a b b c c d d r r . S o , i . e . . 1 2 ( a + b + c + d ) × 2 r = 6 0 0 1 2 × 5 0 × 2 r = 6 0 0 r = 1 2 2 1 . D . N o t e t h a t a l l o f t h e a l t e r n a t i v e s g i v e n a r e o f t h e f o r m s o w e n e e d . T h e o n l y o r d e r e d p a i r s o f p o s i t i v e i n t e g e r s w h i c h s a t i s f y t h i s a r e ( 1 , 6 ) , ( 2 , 3 ) , ( 3 , 2 ) , ( 6 , 1 ) . F o r t h e s e , t h e v a l u e s o f a r e 7 3 , 2 2 , 1 7 , 3 8 r e s p e c t i v e l y . S o t h e r e q u i r e d n u m b e r i s . ( x + y 2 ) 2 = x 2 + 2 x y 2 + 2 y 2 a + 1 2 2 x y = 6 ( x , y ) x 2 + 2 y 2 5 4 + 1 2 2 2 2 . B L e t t h e p e r p e n d i c u l a r f r o m m e e t a t a n d l e t . N o t e t h a t a s a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e r a d i u s a t t h e p o i n t o f c o n t a c t . T h e r e f o r e . C o n s
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So here 1 a 4 y b and 16 4 2 y y c We see that 16 4 4 4 4 2...

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