- #1

chwala

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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: