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Before 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 $\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{7.25}{-0.5}=-14.5$

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 $(\alpha\beta)^3=-0.125$

@Daniel Lim
–
Nope, if the coefficients of a certain quadratic equation are $a,b,c$ and its roots are $\alpha, \beta$, then construct a new equation in terms of $a,b,c$ if the roots of the new equation are

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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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$a,b,c$ and its roots are $\alpha, \beta$, then construct a new equation in terms of $a,b,c$ if the roots of the new equation are

Nope, if the coefficients of a certain quadratic equation arei) $\frac{\alpha}{\beta}, \frac{\beta}{\alpha}$

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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