Question

plz ..solve the ques .5 , 6 .. as soon as possible .. help me frnds .. solve on copy .. n send a pic .. plz

Answer 1

for answer of 1st question see the photo

Answer 2

Hello Dear !
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5) By Division lemma , Find the HCF of 847 and 2160 .Check if they are co prime 
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Solution :
Apply EDL on 847 and 2160 
2160 = 847 X 2 + 466 

Hence r ≠ 0 

So , Again apply EDL on 847 and 466 
847 = 466 X 1 + 381 

Hence r ≠ 0 

So, Again apply EDL on 466 and 381 
466 = 381 X 1 + 85

Hence r ≠ 0 

So , Again apply EDL on 381 and 85 
381 = 85 X 4 + 41 

Hence r ≠ 0 

So, Again apply EDL on 85 and 41 
85 = 41 X 2 +3 

Hence r ≠ 0 

So, Again apply EDL on 41 and 3 
41 = 3 X 13 + 2 

Hence r ≠ 0 

So, Again apply EDL on 3 and 2 
3 = 2 x 1 + 1 

Hence r ≠ 0 

So, Again apply EDL on 2 and 1
2 = 1 x 2 + 0

r = 0

SO the HCF is 1
So these numbers are co - prime
Note : Coprime means couple of two digits which haven't any common factor another than one 
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6) Verify Relationship , After finding the zeros of a (x²+1) - x(a² + 1 ) = 0
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here equation => a (x²+1) - x(a² + 1 ) = 0
Quadratic polynomial = a (x²+1) - x(a² + 1 )
                                   = ax² + a - a²x - x
                                   = ax² - x -a²x + a [ax² -(1+a²)x + a )]
                                   = x(ax-1) -a(ax-1)
                                   = (ax-1) (x-a) 
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So, zeros are x = 1/a and x = a 
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Relation between coefficients :

-b/a = 1/a + a
-(-1 - a²) /a = 1+a² /a
1+a²/a        =  1+a² /a
LHS = RHS 

c/a = 1/a x a 
a/a = a/a 
1    = 1 
LHS = RHS
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