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Find integers $a,b,c,d$ and $e$ such that $\sqrt{3} −\sqrt{5}$ is a solution to the equation: $$ax^4 + bx^3 + cx^2 + dx + e = 0.$$

Being new to quartic equations I wasn't sure how to solve this problem, must we use the quartic formula?

Any help is greatly appreciated!

Thanks

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    Let $x=\sqrt{3}-\sqrt{5}$. Square both sides. What do you have? A square root... isolate it and square again... – 2017-02-20

4 Answers 4

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Since $$\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a),$$ we have that $\sqrt3-\sqrt5$ is a root of the equation $$2(3x^2+5x^2+15)-x^4-9-25=0$$ or $$x^4-16x^2+4=0$$

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$$x=\sqrt{3}-\sqrt{5}$$ $$x^2=8-2\sqrt{15}$$ $$x^2-8=-2\sqrt{15}$$ $$x^4-16x^2+64=60$$ $$x^4-16x^2+4=0$$

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We have $$x =\sqrt {3}-\sqrt {5} $$ $$\Rightarrow x^2 =8-2\sqrt {15} $$ $$\Rightarrow 2\sqrt{15} = 8-x^2$$ $$\Rightarrow 4 (15) = (8-x^2)^2 =x^4-16x^2+64$$ $$\Rightarrow x^4-16x^2+4=0$$

Hope it helps.

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    Just a detail: this isn't a chain of equivalent statements; the last one is strictly weaker than the first. – 2017-02-20
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if we have $$x=\sqrt{3}-\sqrt{5}$$ we get by squaring $$x^2=8-2\sqrt{15}$$ sisolating the square root we obtain $$2\sqrt{15}=8-x^2$$ squaring again we get $$60=64-16x^2+x^4$$ can you finish?