Question

3. If the value of one of the roots of the equation x2 + px = -12 is three times the value of the other,
find the possible value(s) of p.​

Answer 1

Answer:

If one root of the equation x2+px+q=0 is the square of the other, then how do I show that p3−q(3p−1)+q2=0 ?

For quadratic equation ax2+bx+c=0 :

sum of roots =−b/a

product of roots =c/a

If one root of the equation

x2+px+q=0

is the square of the other, then roots are ω and ω2

sum of roots =−p⟹p=−ω2−ω

product of roots =q⟹q=ω2⋅ω=ω3

p3−q(3p−1)+q2

=(−ω2−ω)3−ω3(3(−ω2−ω)−1)+(ω3)2

=(−ω(ω+1))3−ω3(−3ω2−3ω−1)+ω6

=−ω3(ω3+3ω2+3ω+1)+3ω5+3ω4+ω3+ω6

=−ω6−3ω5−3ω4−ω3+3ω5+3ω4+ω3+ω6

=0

Recall which say that p is the opposite of the sum of the roots and q is the product of the roots.

Let a and b be the two roots of the equation, with b=a2 . Then :

p=−(a+b)=−(a+a2)

q=ab=a3

Hence :

p3=−(a3+3a4+3a5+a6) simply by expanding.

−q(3p−1)=−a3(3(−a−a2)−1)=3a4+3a5+a3

q2=a6

So :

p3−q(3p−1)+q2=0

Hope it helped !