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Need urgent help!

Q) How do I go about solving this equation?

I missed a lecture on it and the lecturer hasn't put up any work up
on it yet. :frown:

:frown:

https://gyazo.com/cecacb87537926b92213aca1372d52c3

Could someone explain the method on how to go about solving the question. :puppyeyes:

:puppyeyes:

Original post by XxKingSniprxX

Q) How do I go about solving this equation?

I missed a lecture on it and the lecturer hasn't put up any work up
on it yet. :frown:

:frown:

https://gyazo.com/cecacb87537926b92213aca1372d52c3

Could someone explain the method on how to go about solving the question. :puppyeyes:

:puppyeyes:

Can your studies in Further Maths not help you?

Roots of polynomials???

OP

Original post by Maths is Life

Can your studies in Further Maths not help you?

Roots of polynomials???

I didn't do further maths. :frown:

:frown:

I only self-taught FP1 over the summer and can't remember much of it.

Original post by XxKingSniprxX

I didn't do further maths. :frown:

:frown:

I only self-taught FP1 over the summer and can't remember much of it.

Alpha and beta are the same in each expression.
I cba with the symbols sorry.

Sum if the roots is -b/a
Product of the roots is c/a

It's the morning - go over FP1 roots of polynomials

Hint: for the first expression a=2

For the other equation a is the largest denominator...

(edited 7 years ago)

Study Forum Helper

Original post by XxKingSniprxX

Q) How do I go about solving this equation?

I missed a lecture on it and the lecturer hasn't put up any work up
on it yet. :frown:

:frown:

https://gyazo.com/cecacb87537926b92213aca1372d52c3

Could someone explain the method on how to go about solving the question. :puppyeyes:

:puppyeyes:

In further maths you cover that when a quadratic in the form

ax^2+bx+c=0 \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0

has roots

\alpha

and

\beta

then it must be true that

x^2+\frac{b}{a}x+\frac{c}{a}=(x-\alpha)(x-\beta)

and from expanding RHS you get

x^2+\frac{b}{a}x+\frac{c}{a}=x^2-(\alpha+\beta)x+\alpha \beta

and now by comparison of coefficients you can see that

\frac{b}{a}=-(\alpha+\beta) \Rightarrow -\frac{b}{a}=\alpha+\beta

and

\frac{c}{a}=\alpha \beta

From here you can get your values for

\alpha+\beta

and

\alpha \beta

When it comes to getting a quadratic with roots

\alpha+\frac{1}{\beta}

and

\beta+\frac{1}{\alpha}

you know that:

-q=(\alpha+\frac{1}{\beta})+( \beta + \frac{1}{\alpha})

r=(\alpha+\frac{1}{\beta})(\beta+\frac{1}{\alpha})

and you need to manipulate the RHS on each expression to get it into terms of

\alpha+\beta

and

\alpha \beta

Once you've done that, just substitute the values you found earlier and you got it.

(edited 7 years ago)

KnowledgeIsBest

I feel so dumb, compared to all of you.

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