Value of 'x'

Algebra Level 3

If $x = (x^3 + 6x^2)^\frac{1}{4}$ , find the sum of all the possible solutions of $x$ .

1 3 -1 0

1 solution

A Former Brilliant Member
May 14, 2015

Take the given expression to the 4th power: x^4=x^3+6x^2;

Re-arrange and factor out x^2: x^2(x^2−x−6)=0;

Factorize: x^2(x−3)(x+2)=0;

So, the roots are x=0, x=3 and x=−2. But x cannot be negative as it equals to the even (4th) root of some expression , thus only two solution are valid x=0 and x=3.

The sum of all real possible solutions for x is 0+3=3.

you should write 'all possible real solutions of x.'

lk sharma - 6 years ago

changed thanks

A Former Brilliant Member - 6 years ago

actually the given eqution must have 4 roots and why you neglected -2 i can't understand

Abhishek Sonu - 5 years, 10 months ago

From the given equation you get a polyonyme, which should have 4 roots. It does, 0 as a double root, -2 and 3. We produced the polyonyme from the equation, by applying both ends to the function f(x)=x^4. This is not an invertible function when x is real, so we can't get the expression from the polyonyme, so not all roots of it will necessarily solve the equation. More simply, if E is our given equation:

x^4=x^3+6x^2<=>+-E ---->True

E=>x^4=x^3+6x^2 ----> True

x^4=x^3+6x^2=>E ----> False

Nick Zafiridis - 5 years, 6 months ago

Value of 'x'

1 solution

1 pending report

Vote up reports you agree with