If p and q are the roots of the equation x2-7x+10=0 ,the value of p/q-p/q=
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Solution :
Given p and q are the roots of
the equation x² - 7x + 10 = 0
compare above equation with
ax² + bx + c = 0 , we get
a = 1 , b = -7 , c = 10
i ) sum of the roots = -b/a
=> p + q = - (-7)/1 = -7
ii ) product of the roots = c/a
=> pq = 10/1 = 10
iii ) ( p - q )² = ( p + q )² - 4pq
= ( -7 )² - 4×10
= 49 - 40
= 9
p - q = ± √9
=> p - q = 3
Now ,
p/q - q/p = ( p² - q² )/pq
= [( p + q )( p - q )]/pq
= [ ( -7 ) 3 ]/10
= -21/10
••••
Given p and q are the roots of
the equation x² - 7x + 10 = 0
compare above equation with
ax² + bx + c = 0 , we get
a = 1 , b = -7 , c = 10
i ) sum of the roots = -b/a
=> p + q = - (-7)/1 = -7
ii ) product of the roots = c/a
=> pq = 10/1 = 10
iii ) ( p - q )² = ( p + q )² - 4pq
= ( -7 )² - 4×10
= 49 - 40
= 9
p - q = ± √9
=> p - q = 3
Now ,
p/q - q/p = ( p² - q² )/pq
= [( p + q )( p - q )]/pq
= [ ( -7 ) 3 ]/10
= -21/10
••••
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