- #1
chwala
Gold Member
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- Homework Statement
- see attached
- Relevant Equations
- quadratic equations
I am refreshing on this...Have to read broadly...i will start with (b) then i may be interested in alternative approach or any correction that may arise from my working. Cheers.
Kindly note that i do not have the solutions to the following questions...
For (b), we know that,
say, if ##x=α## and ##x=β## are roots of the given quadratic equation, then it follows that,
##-(α+β)=3.5##
##αβ=2##,
##(α-β)^2=(α+β)^2-4αβ##
##(α-β)^2=(3.5)^2-8##
##(α-β)^2=\dfrac{17}{4}##
For (c),
##α^3+β^3=(α+β)(α^2-αβ+β^2)##
##α^2+β^2=(α+β)^2-2αβ##
##α^2+β^2=(3.5)^2-4##
##α^2+β^2=\dfrac{33}{4}##
Therefore,
##α^3+β^3=(α+β)(α^2+β^2-αβ)##
##α^3+β^3=(-3.5)(8.25-2)##
##α^3+β^3=(-3.5)(6.25)##
##α^3+β^3=-\dfrac{175}{8}##
For part (d),
##α^3-β^3=\sqrt{\dfrac {17}{4}}⋅\dfrac {41}{4}=\dfrac {41}{4}⋅{\dfrac {\sqrt17}{2}}=\dfrac {41}{8}\sqrt 17##
For part (e),
We shall have,
##x^2-\dfrac{α^3+β^3}{αβ}x+αβ=0##
##x^2-\dfrac{175}{16}x+2=0##
##⇒16x^2-175x+32=0##
Bingo!
Kindly note that i do not have the solutions to the following questions...
For (b), we know that,
say, if ##x=α## and ##x=β## are roots of the given quadratic equation, then it follows that,
##-(α+β)=3.5##
##αβ=2##,
##(α-β)^2=(α+β)^2-4αβ##
##(α-β)^2=(3.5)^2-8##
##(α-β)^2=\dfrac{17}{4}##
For (c),
##α^3+β^3=(α+β)(α^2-αβ+β^2)##
##α^2+β^2=(α+β)^2-2αβ##
##α^2+β^2=(3.5)^2-4##
##α^2+β^2=\dfrac{33}{4}##
Therefore,
##α^3+β^3=(α+β)(α^2+β^2-αβ)##
##α^3+β^3=(-3.5)(8.25-2)##
##α^3+β^3=(-3.5)(6.25)##
##α^3+β^3=-\dfrac{175}{8}##
For part (d),
##α^3-β^3=\sqrt{\dfrac {17}{4}}⋅\dfrac {41}{4}=\dfrac {41}{4}⋅{\dfrac {\sqrt17}{2}}=\dfrac {41}{8}\sqrt 17##
For part (e),
We shall have,
##x^2-\dfrac{α^3+β^3}{αβ}x+αβ=0##
##x^2-\dfrac{175}{16}x+2=0##
##⇒16x^2-175x+32=0##
Bingo!
Last edited:
