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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Top NewestBefore 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. – Daniel Lim · 2 years, 12 months ago
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Nice solution! By the way, here's a challenge for you: can you generalise this to \(f(x)=ax^2+bx+c\)?:) – 敬全 钟 · 2 years, 12 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),
\[f(x)=x^2- \left(\frac{c}{a}\times\frac{-b}{a}\right)x+\left(\frac{c}{a}\right)^3\] – Daniel Lim · 2 years, 12 months ago
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Like what? \(f(x)=2x^2+29x+2\) and \(f(x)=8x^2+10x-1\)? – Daniel Lim · 2 years, 12 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) \(\alpha^2\beta, \alpha\beta^2\) – 敬全 钟 · 2 years, 12 months ago
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Ok, I'll try – Daniel Lim · 2 years, 12 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 . – Shriram Lokhande · 2 years, 12 months ago
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oh, that's right, I will edit it
Thanks ;D – Daniel Lim · 2 years, 12 months ago
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Great :D – Victor Loh · 2 years, 12 months ago
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Thanks – Daniel Lim · 2 years, 12 months ago
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