Solving a quadratic equation as a sum and product of its roots

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  • #26
chwala
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so i was just right after all....take your vote:smile:
 
  • #27
haruspex
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##\frac {1}{∝^3}##+##\frac {1}{β^3}##=##\frac {-b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\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##
If the sum of the roots is ##\frac {-b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}## then the coefficient of x in the equation will be ##\frac {+b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}##
 
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  • #28
chwala
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If the sum of the roots is ##\frac {-b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}## then the coefficient of x in the equation will be ##\frac {+b}{a}##. ##[\frac{b^2-3ac}{a^2}]##. ##\frac {a^3}{c^3}##
yeah, now its very clear to me...just small detail that i had missed thanks Haruspex.
 
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  • #29
chwala
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1593582929694.png


This was the second part of the question and i have used this approach below;

1593583056745.png
 
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  • #30
chwala
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The approach below is from a colleague, let me know of any other approach to this part of the question.

1593583162924.png
 
  • #31
PeroK
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The approach below is from a colleague, let me know of any other approach to this part of the question.

View attachment 265589
Your solution is a lot neater than that.
 
  • #32
chwala
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Thanks Perok, it took me a while to think on that...all the passion for this great subject...its a pleasure. I will be posting and sharing more questions...bingo!
 
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