TAKE: \(x=(x^3+6x^2)^\frac{1}{4}\)

Raise both sides to the power of 4 to get: \(x^4=x^3+6x^2\)

Set equation equal to zero: \(x^4-x^3-6x^2=0\)

Factor to get: \(x^2(x^2-x-6)=0\)

Factor part in brackets: \(x^2(x-3)(x+2)=0\)

Solutions: x = 0, x = 3 and x = -2
HOWEVER, we should check these solutions for extraneous roots by plugging the values into the original equation.

Test \(x = 0\)
We get: \(0=(0^3+6(0)^2)^\frac{1}{4}\)
Simplify: \(0=(0)^\frac{1}{4}\)
WORKS!

Test \(x = 3\)
We get: \(3=(3^3+6(3)^2)^\frac{1}{4}\)
Simplify: \(3=(81)^\frac{1}{4}\)
WORKS!

Test \(x = -2\)
We get: \(-2=((-2)^3+6(-2)^2)^\frac{1}{4}\)
Simplify: \(-2=(16)^\frac{1}{4}\)
NO GOOD!

So, the solutions are \(x = 0\) and \(x = 3\)

We get:
Quantity A: \(0+3=3\)
Quantity B: \(1\)

Answer: A

Cheers,
Brent

Sir, do i need to check in original equation? If i put x=-2 in \(x^4=x^3+6x^2\), I get the equal no in both side, Is this ambiguous question?

Doesn't make sense. Coz,

\((16)^\frac{1}{4} = 2\) or \(-2\)

So, \(-2\) checks out.

The summation of the possible values of \(x\) is \(0 + 3 - 2 = 1\).

The answer should be C.

\((16)^\frac{1}{4} = \sqrt[4]{16}\), and the fourth root notation tells us to take only the POSITIVE fourth root of 16.

This is very similar to the square root notation that tells us to take the positive root of a number.
For example, \(\sqrt{9}=3\) (and not -3)

If, however, we're given a similar equation but WITHOUT the root notation, there will be a positive and a negative solution.
For example, if \(x^4 = 81\), then there are two solutions: \(x = 3\) and \(x = -3\)

When testing for extraneous roots, we must plug the potential solution into the original equation

In this case, the original equation is \(x=(x^3+6x^2)^\frac{1}{4}\).

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