Help Pls with 4U Problem ASAP (1 Viewer)

Modern4DaBois · Dec 1, 2021

Hello, could anyone pls help me do part (c)?
I've done a and b and kinda mind-blanked for c.

Screen Shot 2021-12-01 at 11.56.26 am.png

quickoats · Dec 1, 2021

I'm assuming you got somewhere near the required expression but not quite? Hint: consider the roots-coefficient relationship.

Modern4DaBois · Dec 1, 2021

quickoats said:
I'm assuming you got somewhere near the required expression but not quite? Hint: consider the roots-coefficient relationship.

Hmm... I'll give this a try. Thanks!

CM_Tutor · Dec 2, 2021

Some help for (a) in a related thread.

Having established that the solutions are:

$x^2 = \cfrac{-A \pm \sqrt{A^2 - 4B}}{2} = - \cfrac{A \mp \sqrt{A^2 - 4B}}{2} < 0$

there are two pairs of solutions, $x^2 = -c^2$ and $x^2 = -d^2$ where one of $c^2$ is one of

$\cfrac{A + \sqrt{A^2 - 4B}}{2} \qquad \text{or} \qquad \cfrac{A - \sqrt{A^2 - 4B}}{2}$

and $d^2$ is the other and both $c^2$ and $d^2$ are poitive and real.

It follows that

$\begin{align*} c^4 + d^4 &= \left(\cfrac{A + \sqrt{A^2 - 4B}}{2}\right)^2 + \left(\cfrac{A - \sqrt{A^2 - 4B}}{2}\right)^2 \\ &= \cfrac{1}{4}\left[\big(A^2 + 2A\sqrt{A^2 - 4B} + A^2 - 4B\big) + \big(A^2 - 2A\sqrt{A^2 - 4B} + A^2 - 4B\big)\right] \\ &= \cfrac{1}{4}\left(2A^2 + 2A^2 - 2 \times 4B\right) \\ &= \cfrac{4A^2}{4} - \cfrac{8B}{4} \\ &= A^2 - 2B \end{align*}$

Modern4DaBois · Dec 2, 2021

CM_Tutor said:
Some help for (a) in a related thread.

Having established that the solutions are:

$x^2 = \cfrac{-A \pm \sqrt{A^2 - 4B}}{2} = - \cfrac{A \mp \sqrt{A^2 - 4B}}{2} < 0$

there are two pairs of solutions, $x^2 = -c^2$ and $x^2 = -d^2$ where one of $c^2$ is one of

$\cfrac{A + \sqrt{A^2 - 4B}}{2} \qquad \text{or} \qquad \cfrac{A - \sqrt{A^2 - 4B}}{2}$

and $d^2$ is the other and both $c^2$ and $d^2$ are poitive and real.

It follows that

$\begin{align*} c^4 + d^4 &= \left(\cfrac{A + \sqrt{A^2 - 4B}}{2}\right)^2 + \left(\cfrac{A - \sqrt{A^2 - 4B}}{2}\right)^2 \\ &= \cfrac{1}{4}\left[\big(A^2 + 2A\sqrt{A^2 - 4B} + A^2 - 4B\big) + \big(A^2 - 2A\sqrt{A^2 - 4B} + A^2 - 4B\big)\right] \\ &= \cfrac{1}{4}\left(2A^2 + 2A^2 - 2 \times 4B\right) \\ &= \cfrac{4A^2}{4} - \cfrac{8B}{4} \\ &= A^2 - 2B \end{align*}$

Thank you so much bro! That makes sense

Help Pls with 4U Problem ASAP (1 Viewer)

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