Roots Question Help (1 Viewer)

[Jojofelyx]

How does one solve this?

 

[ExtremelyBoredUser]

A) So these are the roots; A, B, A+B:

A+B+A+B =-a/1, as by sum of roots
2(A+B) = -a
A+B = -a/2

and since A+B is a root by itself, x = -a/2

B)

AB+ A(A+B) + B(A+B) = 0, sum of double roots
AB(A+B) = -1 (Triple Roots)
(AB)= -1/(A+B) = -1/(-a/2) = 2/a, pretty much take the negative reciprocal of (A+B) root.

AB + (A+B)(A+B) = 0, through factorising
2/a + (-a/2)^2 =0
2/a + a^2/4 = 0
2 + a^3/4 = 0, multiplying a by both sides

8 + a^3 = 0
a^3 = -8
a = -2

 

[Jojofelyx]

A) So these are the roots; A, B, A+B:

A+B+A+B =-a/1, as by sum of roots
2(A+B) = -a
A+B = -a/2

and since A+B is a root by itself, x = -a/2

B)

AB+ A(A+B) + B(A+B) = 0, sum of double roots
AB(A+B) = -1 (Triple Roots)
(AB)= -1/(A+B) = -1/(-a/2) = 2/a, pretty much take the negative reciprocal of (A+B) root.

AB + (A+B)(A+B) = 0, through factorising
2/a + (-a/2)^2 =0
2/a + a^2/4 = 0
2 + a^3/4 = 0, multiplying a by both sides

8 + a^3 = 0
a^3 = -8
a = -2

Thanks heaps !

 

[CM_Tutor]

Having established that is a root of the equation , you can also substitute:

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[ExtremelyBoredUser]

Having established that is a root of the equation , you can also substitute:

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Ah that's actually really neat because you can save quite a bit of working and its much quicker so you wouldn't make mistakes. Thanks for that.