TarPhi wrote:
Bunuel wrote:
7. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x?
I. -50
II. 25
III. 50
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).
So, we have that \(x=5\), \(x=-5\), \(x=2\), or \(x=-2\).
\(-50=5*(-5)*2\);
\(50=5*(-5)*(-2)\).
Only 25 is NOT a product of three possible values of x
Answer: B.
In this question -
1. If x=\sqrt[4]{x^3+6x^2}, then the sum of all possible solutions for x is:
A. -2
B. 0
C. 1
D. 3
E. 5
Solution: new-algebra-set-149349-60.html#p1200948
x^4=x^3 + 6x^2
x^4 - x^3 - 6x^2 = 0 ------------> x^2 (x^2 - x - 6) = 0 -----------> x^2 (x-3)(x+2) = 0
x=0 or 3 or -2. With x = -2 original equation does not hold true so possible values for X are 3 and 0 Hence their sum is 3
Choice D.
You didn't take the negative values - why are we considering negative values here for 4th roots?
This is explained several times on previous pages.
\(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)).
When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.
That is, \(\sqrt{25}=5\), NOT +5 or -5.
Even roots have only a positive value on the GMAT.In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
.
Even roots (sq root and 4th root) cannot have a -ve value but x^2=16 and x^4=16 can have (+2 or-2) as the solutions. I got it