- #1

chwala

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## Homework Statement:

- If ##∝## and ##β## are the roots of the equation ##ax^2+bx+c=0##, obtain the equation whose roots are ##\frac {1}{∝^3}## and ##\frac {1}{β^3}##

## Relevant Equations:

- quadratic equations

for the sum,

##\frac {1}{∝^3}##+##\frac {1}{β^3}##=##\frac {β^3+∝^3}{∝^3β^3}##

=##\frac {(∝+β)[(∝+β)^2-3∝β]}{∝^3β^3}##

=##\frac {-b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}##

for the product,

##\frac {1}{∝^3}## . ##\frac {1}{β^3}##=##\frac {a^3}{c^3}##

therefore, on multiplying each term by ##c^3##, the equation takes the form,

##c^3x^2-b(b^2-3ac)x+a^3=0##

now where my problem is that the textbook indicates that the answer is

##c^3x^2+b(b^2-3ac)x+a^3=0## this is where i need clarification. In my understanding, i.e in reference to sum of roots

where if ##m## and ##n## are roots of ##ax^2+bx+c=0##, then it follows that ##m+n##=##\frac {-b}{a}##

##\frac {1}{∝^3}##+##\frac {1}{β^3}##=##\frac {β^3+∝^3}{∝^3β^3}##

=##\frac {(∝+β)[(∝+β)^2-3∝β]}{∝^3β^3}##

=##\frac {-b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}##

for the product,

##\frac {1}{∝^3}## . ##\frac {1}{β^3}##=##\frac {a^3}{c^3}##

therefore, on multiplying each term by ##c^3##, the equation takes the form,

##c^3x^2-b(b^2-3ac)x+a^3=0##

now where my problem is that the textbook indicates that the answer is

##c^3x^2+b(b^2-3ac)x+a^3=0## this is where i need clarification. In my understanding, i.e in reference to sum of roots

where if ##m## and ##n## are roots of ##ax^2+bx+c=0##, then it follows that ##m+n##=##\frac {-b}{a}##

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