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# Problem 2: Symmetrical Properties of Roots

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) $$\alpha^2\beta$$, $$\alpha\beta^2$$

Note by Victor Loh
2 years, 8 months ago

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

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. · 2 years, 8 months ago

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Nice solution! By the way, here's a challenge for you: can you generalise this to $$f(x)=ax^2+bx+c$$?:) · 2 years, 8 months ago

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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),

$f(x)=x^2- \left(\frac{c}{a}\times\frac{-b}{a}\right)x+\left(\frac{c}{a}\right)^3$ · 2 years, 8 months ago

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Like what? $$f(x)=2x^2+29x+2$$ and $$f(x)=8x^2+10x-1$$? · 2 years, 8 months ago

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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

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

ii) $$\alpha^2\beta, \alpha\beta^2$$ · 2 years, 8 months ago

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Ok, I'll try · 2 years, 8 months ago

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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 . · 2 years, 8 months ago

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oh, that's right, I will edit it

Thanks ;D · 2 years, 8 months ago

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Great :D · 2 years, 8 months ago

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Thanks · 2 years, 8 months ago

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