roots of polynomials (1 Viewer)

[Smilebuffalo]

two roots of x^3 + mx^2 + 15x - 7 = 0 are equal and rational. Find m.

?

answer: m = -9

 

[tommykins]

@, $ be roots.

@^2.$ = 7 -> $ = 7/@^2

@ + @ + $ = -m

@^2 + @$ + @$ = 15 -> @^2 + 2@$ = 15 -> @^2 + 2@.7/@^2 = 15
@^2 + 14/@ = 15
@^3 + 14 = 15@
@^3 - 15@ + 14 = 0

Solving that, you get @ as 1 (since @ is rational)

hence 1.$ = 7, $ = 7

@ + @ + $ = -m
1+1+7 = -m
m = -9

probably a better way but yeah.

 

[Trebla]

You could also use a theorem from Ext2 that a double root satisfies P(x) = P'(x) = 0

 

[Smilebuffalo]

ahh i see. So its necessary to solve a cubic equation in order to solve this problem?

how did you solve the cubic: @^3 - 15@ + 14 = 0 ???

 

[tommykins]

observation. first number to sub in is +-1