Problem 2: Symmetrical Properties of Roots

If α\alpha and β\beta are the roots of the equation 2x25x1=02x^2-5x-1=0, form equations whose roots are

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

(b) α2β\alpha^2\beta, αβ2\alpha\beta^2

Note by Victor Loh
4 years, 11 months ago

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

α+β=2.5\alpha+\beta=2.5

αβ=0.5\alpha\beta=-0.5

Solution (a):

The equation should be in the form x2(αβ+βα)x+1x^2-(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha})x+1

We can rewrite αβ+βα\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} to be α2+β2αβ\dfrac{\alpha^2+\beta^2}{\alpha\beta}. α2+β2\alpha^2+\beta^2 is not hard to find, as it is(α+β)22αβ=7.25(\alpha+\beta)^2-2\alpha\beta=7.25. Then we can easily get α2+β2αβ=7.250.5=14.5\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{7.25}{-0.5}=-14.5

Therefore, (xαβ)(xβα)=x2+14.5x+1(x-\dfrac{\alpha}{\beta})(x-\dfrac{\beta}{\alpha})=\boxed{x^2+14.5x+1}

Solution (b):

The equation should be in the form x2(α2β+αβ2)x+α3β3x^2-(\alpha^2\beta+\alpha\beta^2)x+\alpha^3\beta^3

To find α2β+αβ2\alpha^2\beta+\alpha\beta^2, we can factorize it to be αβ(α+β)\alpha\beta(\alpha+\beta). It's lot more easier now, and we get α2β+αβ2=1.25\alpha^2\beta+\alpha\beta^2=-1.25

Now we need to find α3β3\alpha^3\beta^3, which is not so difficult as we can write it as (αβ)3=0.125(\alpha\beta)^3=-0.125

Therefore, (xα2β)(xαβ2)=x2+1.25x0.125(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 2x2+29x+22x^2+29x+2 while the equation for (b) can be written as 8x2+10x18x^2+10x-1 as it will look better.

Daniel Lim - 4 years, 11 months ago

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Great :D

Victor Loh - 4 years, 11 months ago

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Thanks

Daniel Lim - 4 years, 11 months ago

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you could rewrite the solution for (a) to be 2x2+29x+22x^2+29x+2 & of (b) to be 8x2+10x18x^2+10x-1. integers seem better in equations than fractions .

Shriram Lokhande - 4 years, 11 months ago

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oh, that's right, I will edit it

Thanks ;D

Daniel Lim - 4 years, 11 months ago

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Nice solution! By the way, here's a challenge for you: can you generalise this to f(x)=ax2+bx+cf(x)=ax^2+bx+c?:)

敬全 钟 - 4 years, 11 months ago

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Like what? f(x)=2x2+29x+2f(x)=2x^2+29x+2 and f(x)=8x2+10x1f(x)=8x^2+10x-1?

Daniel Lim - 4 years, 11 months ago

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

i) αβ,βα\frac{\alpha}{\beta}, \frac{\beta}{\alpha}

ii) α2β,αβ2\alpha^2\beta, \alpha\beta^2

敬全 钟 - 4 years, 11 months ago

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@敬全 钟 Ok, I'll try

Daniel Lim - 4 years, 11 months ago

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For (a),

f(x)=x2((ba)22caca)x+1f(x)=x^2-\left(\frac{\left(\frac{-b}{a}\right)^2-\frac{2c}{a}}{\frac{c}{a}}\right)x+1

For (b),

f(x)=x2(ca×ba)x+(ca)3f(x)=x^2- \left(\frac{c}{a}\times\frac{-b}{a}\right)x+\left(\frac{c}{a}\right)^3

Daniel Lim - 4 years, 11 months ago

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