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What is the sum of the roots of the equation listed below?
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\[(x-\sqrt2)(x^2-\sqrt3x+\pi)=0\]
for a quadratic \(ax^2+bx+c\) the sum of the roots is \(-b/a\); observe: $$a(x-r_1)(x-r_2)=a(x^2-(r_1+r_2)x+r_1r_2)=ax^2\underbrace{-a(r_1+r_2)}_bx+\underbrace{ar_1r_2}_c\\\implies b =-a(r_1+r_2)\\\implies r_1+r_2=-\frac{b}a$$
so the sum of the roots of \(x^2-\sqrt3x+\pi\) is \(-(-\sqrt3)/1=\sqrt3\), and then the root of \(x-\sqrt2\) is clearly \(\sqrt2\) so the total sum of our roots is \(\sqrt2+\sqrt3\). this is because the roots of a product of two polynomials is simply the combined roots of both individual polynomials
i see, thanks
Sum of the Roots of a Polynomial Theorem Attached
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this chart is really helpful, thanks
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