if alpha and beta are the zeros of fx= ax²+bx+c then evaluate alpha²/beta² + beta²/alpha². as shown in the picture.
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Answer:
Step-by-step explanation:
Roots of this quadratic eqn are
r1=-b/2a and r2=c/a
r1²/r2² + r2²/r1²
(-b/2a)²/(c/a)² + (c/a)²/(-b/2a)²
=b²/4c² + 4c²/b²
=(b/2c)² + (2c/b)²
lucifer96:
it is not ful answer
I don't have the values....u coloured Dem
that is not a value
you don't no how do this
Let the zeros be s and t
Sorry the previous one was incorrect....but this is correct
Answered by
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Answer:
Step-by-step explanation:
alpha²/beta²+beta²/alpha²
By taking LCM..
alpha⁴+beta⁴/alpha²beta²
[ Here, a⁴+b⁴= (a+b)⁴ - 4ab(a+b)² + 2(ab)² ]
So we have,
(alpha+beta)⁴ - 4(alpha x beta)(alpha+beta)² + 2(alpha x beta)² / (alpha x beta)²
= (-b/a)⁴ - 4(c/a)(-b/a)² + 2(c/a)² / (c/a)²
= b⁴/a⁴ - 4b²c/a³ + 2c²/a² / (c²/a²)
Taking LCM of above...
= b⁴ - 4ab²c + 2a²c² / a⁴ / c²/a²
= b⁴ - 4ab²c + 2a²c² / a²c² (Answer)..
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