Polynomials 2 (1 Viewer)

coyazayo · Mar 5, 2016

Hai ya everyone reading this thread. As always thanks in advance

The cubic equation x³+px²+qx+30=0, where p and q are rational numbers, has one root x=1+√2

i) Find the other roots of the equation
ii) Hence, or otherwise, find the values of p and q

KingOfActing · Mar 5, 2016

coyazayo said:
Hai ya everyone reading this thread. As always thanks in advance

The cubic equation x³+px²+qx+30=0, where p and q are rational numbers, has one root x=1+√2

i) Find the other roots of the equation
ii) Hence, or otherwise, find the values of p and q

$Since the coefficients are rational, we know that the rational conjugate of the given root will also be a root:$\\x^3+px^2+qx+30 = (x-\alpha)(x-1-\sqrt2)(x-1+\sqrt2) = (x-\alpha)(x^2-2x-1)\\$Comparing the constant term (or substituting $x=0$) we obtain $\alpha = 30\\$Thus, the roots are $ 1+\sqrt2, 1-\sqrt2, 30$\\To obtain the values of $p$ and $q$ simply expand out the equation from the part before and compare coefficients.$

The other way to find p and q if you're interested is to simply substitute x = 1+√2 into the original equation, then equate rational and irrational parts.

Polynomials 2 (1 Viewer)

coyazayo

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KingOfActing

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