If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^2-5x-1=0\), form equations whose roots are

(a) \(\frac{\alpha}{\beta}\), \(\frac{\beta}{\alpha}\)

(b) \(\alpha^2\beta\), \(\alpha\beta^2\)

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## Comments

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TopNewestBefore attempting to solve these, it is better to find out what \(\alpha+\beta\) and \(\alpha\beta\) is as we can easily see that we can try and use the Vieta's Formula to solve these.

\(\alpha+\beta=2.5\)

\(\alpha\beta=-0.5\)

Solution (a):

The equation should be in the form \(x^2-(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha})x+1\)

We can rewrite \(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\) to be \(\dfrac{\alpha^2+\beta^2}{\alpha\beta}\). \(\alpha^2+\beta^2\) is not hard to find, as it is\((\alpha+\beta)^2-2\alpha\beta=7.25\). Then we can easily get \(\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{7.25}{-0.5}=-14.5\)

Therefore, \((x-\dfrac{\alpha}{\beta})(x-\dfrac{\beta}{\alpha})=\boxed{x^2+14.5x+1}\)

Solution (b):

The equation should be in the form \(x^2-(\alpha^2\beta+\alpha\beta^2)x+\alpha^3\beta^3\)

To find \(\alpha^2\beta+\alpha\beta^2\), we can factorize it to be \(\alpha\beta(\alpha+\beta)\). It's lot more easier now, and we get \(\alpha^2\beta+\alpha\beta^2=-1.25\)

Now we need to find \(\alpha^3\beta^3\), which is not so difficult as we can write it as \((\alpha\beta)^3=-0.125\)

Therefore, \((x-\alpha^2\beta)(x-\alpha\beta^2)=\boxed{x^2+1.25x-0.125}\)

Edited:The equation for (a) can also be written as \(2x^2+29x+2\) while the equation for (b) can be written as \(8x^2+10x-1\) as it will look better.

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Nice solution! By the way, here's a challenge for you: can you generalise this to \(f(x)=ax^2+bx+c\)?:)

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For (a),

\[f(x)=x^2-\left(\frac{\left(\frac{-b}{a}\right)^2-\frac{2c}{a}}{\frac{c}{a}}\right)x+1\]

For (b),

\[f(x)=x^2- \left(\frac{c}{a}\times\frac{-b}{a}\right)x+\left(\frac{c}{a}\right)^3\]

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Like what? \(f(x)=2x^2+29x+2\) and \(f(x)=8x^2+10x-1\)?

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i) \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha}\)

ii) \(\alpha^2\beta, \alpha\beta^2\)

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you could rewrite the solution for

(a)to be \(2x^2+29x+2\) & of(b)to be \(8x^2+10x-1\). integers seem better in equations than fractions .Log in to reply

oh, that's right, I will edit it

Thanks ;D

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Great :D

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Thanks

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