Ergenekon wrote:
Bunuel wrote:
SOLUTIONs:
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
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 (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).
The sum of all possible solutions for x is 0+3=3.
Answer: D.
Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power.
From GMATCLUB math book:
General rules:
• \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\).
• \((\sqrt{x})^n=\sqrt{x^n}\)
• \(x^{\frac{1}{n}}=\sqrt[n]{x}\)
• \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\)
• \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\)
• \(\sqrt{x^2}=|x|\), when \(x\leq{0}\), then \(\sqrt{x^2}=-x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)
• 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. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
Even roots have only a positive value on the GMAT.• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
More on this can be found here:
math-number-theory-88376.html#p666609 _________________
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