Polynomial Question (1 Viewer)

[M@ster P]

Hi i need help with this question:

a, b, ab are the roots of the equation 4x^3 + 6x^2 + c = 0, where c is a non-zero constant.

i. show that ab =/= 0
ii. Show that ab + a^2b + ab^2 = 0 and deduce the value of a + b
iii. Show that ab = -0.5
iv. Solve the equation to determine the value of c

 

[tommykins]

i. a.b.ab = a²b² = -c/4
ab = -sqrt(c)/2 as c =/= 0, ab =/= 0.

ii. Sum of roots two at a time = coefficient of x/4 = 0/4 = 0.
Thus ab + a²b + ab² = 0.
ab(1+a+b) = 0
1+a+b = 0
Thus a+b = -1

iii. Sum of roots = -6/4 = -3/2
a+b+ab = -3/2
-1 + ab = -3/2
ab = -1/2

iv. Seeing as ab = -1/2, product of roots is a.b.ab = a²b² = -c/4
thus 1/4 = -c/4
1 = -c
Thus c = -1

 

[M@ster P]

sorry but I still don't understand part 1, can someone mind explaining it for me

 

[tommykins]

The only way to get ab by itself is by product of roots as we don't know the value of a+b.

Doing products of the roots, you get a*b*ab = a²b² = -constant/a = -c/4
Square rooting both sides, you get ab = sqrt(-c)/2
As we know c is not equal to zero, then sqrt(-c) is not equal to 0, hence ab =/= 0.

 

[M@ster P]

ok you can say c cannot equal to 0 or infact any number since you cannot square root a negative number. And since this is true you can then say ab =/= 0.

Is this right?

 

[tommykins]

ok you can say c cannot equal to 0 or infact any number since you cannot square root a negative number. And since this is true you can then say ab =/= 0.

Is this right?

Not exactly.

a, b, ab are the roots of the equation 4x^3 + 6x^2 + c = 0, where c is a non-zero constant.

It is any number that is not 0, we don't know if it's positive or negative yet.

Further working shows that c = -1, hence sqrt(-c) = sqrt(-[-1]) = sqrt1 = 1.

Works out fine.

 

[M@ster P]

why do you assume c =/= to 0, is it because of the question?

 

[tommykins]

The question says that c is a NON-ZERO constant, meaning c =/= 0.

 

[M@ster P]

ok thanks i get it now, i guess i missed an important part of the question