If alpha and beta are the zeroes of the quadratic polynomial p(x)=ax2+bx+c then evaluate (1/alpha-1/beta)
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Answered by
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P(x) = ax² + bx + c
By Root-Coefficient relation,
A + B = (-b/a)
AB = (c/a)
=> (A-B)² = A² + B² - 2AB = A² + B² +2AB - 2AB - 2AB
=> (A-B)² = (A+B)² - 4AB
Substituting values in the eqn.
=> ( A-B )² = (b²/a²) - 4c/a = ( b² - 4ac ) / a²
=> ( A-B ) = ±√(b² - 4ac) / a
Now, 1/A - 1/B = -( A - B ) / AB = -(±√(b² - 4ac) / a) / (c/a)
= -√(b² - 4ac) / c or +√(b² - 4ac)
By Root-Coefficient relation,
A + B = (-b/a)
AB = (c/a)
=> (A-B)² = A² + B² - 2AB = A² + B² +2AB - 2AB - 2AB
=> (A-B)² = (A+B)² - 4AB
Substituting values in the eqn.
=> ( A-B )² = (b²/a²) - 4c/a = ( b² - 4ac ) / a²
=> ( A-B ) = ±√(b² - 4ac) / a
Now, 1/A - 1/B = -( A - B ) / AB = -(±√(b² - 4ac) / a) / (c/a)
= -√(b² - 4ac) / c or +√(b² - 4ac)
modanwal:
not explained in proper way
K Bro...... Thx for thy SUGGESTION...
Further need of Explanation would suggest you're a beginner at Mathematics........ HEHE...... But if you insist.
I'll try
Answered by
1
Sum of zeroes=alpha+beta=-b/a. Product of zeroes=alpha*beta=c/a
(alpha-beta)square =(alpha+beta)sq-4*alpha*beta
=(-b/a)sq-4*c/a =(b)sq/(a)sq-4c/a = ((b)sq-4ac)/(a)sq
(aplha*beta)sq=(c)sq/(a)sq
(1/alpha-1/beta)sq=(beta-alpha)sq/(alpha*beta)sq= ((b)sq-4ac)/(c)sq
(1/alpha-1/beta)= sq root ((b)sq-4ac)/c
(alpha-beta)square =(alpha+beta)sq-4*alpha*beta
=(-b/a)sq-4*c/a =(b)sq/(a)sq-4c/a = ((b)sq-4ac)/(a)sq
(aplha*beta)sq=(c)sq/(a)sq
(1/alpha-1/beta)sq=(beta-alpha)sq/(alpha*beta)sq= ((b)sq-4ac)/(c)sq
(1/alpha-1/beta)= sq root ((b)sq-4ac)/c
HEHE! Seems more like a JAVA Coding to me........ Use the Symbols Buddy
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