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IMO 1959 Q3

Hi!I am currently training for the mathematical olympiad,and I am stuck at the question below. Please help by writing a solution for me,using simple to understand terms.Thanks!

Let \(a, b, c\) be real numbers. Given the equation for \(\cos x\): \[a \cos^2(x )+ b \cos (x) + c = 0\] Form a quadratic equation in \(\cos (2x)\) whose roots are the same values of \(x\). Compare the equations in \(\cos x\) and \(\cos (2x)\) for \(a=4, b=2, c=-1\).

Note by Timothy Wan
2 years ago

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Hi, since this question is pretty easy I'll just give you the link for the solution to the given problem which is pretty much the same method used by most people .And if you still don't get it , reply here . Azhaghu Roopesh M · 2 years ago

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@Azhaghu Roopesh M I do not get how they formed the quadratic equation.Can you help to explain? Timothy Wan · 2 years ago

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@Timothy Wan Well you see any quadratic equation in any variable (let's take x) can be written as [ \( x^{2} \) - (Sum of roots)\(x\) + (Product of roots) ] .So they found out the sum and product of the roots of the quadratic(quadratic in \( \cos x \) in your case ) and then they just rearranged it .Now you are given the values of a , b and c so substitute them to get your final quadratic equation in \( \cos x \) .

Now just substitute the values of a , b and c in the given equation and compare the coefficients of both the quadratic equation .

Hope I was of some help . Azhaghu Roopesh M · 2 years ago

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@Azhaghu Roopesh M Yes!Thank you! Timothy Wan · 2 years ago

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Multiply by 2,2cos^2x will become 1+cos2x, then take the 2bcosx to the right and square both sides Aviroop Pal · 2 years ago

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