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21 Jun 2014, 10:37
Kudos
Bookmarks
Ergenekon
Bunuel
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.
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
06 Aug 2014, 11:00
Kudos
Bookmarks
vik09
The equation x^2+ax−b=0 has equal roots, and one of the roots of the equation x^2+ax+15=0 is 3. What is the value of b?
a)−64 b) −16 c) −15 d) −116 e) −164
Dear vik09,a)−64 b) −16 c) −15 d) −116 e) −164
This is a tricky question, and I am happy to help.
First of all, you may find these blogs helpful:
https://magoosh.com/gmat/2012/foil-on-th ... expanding/
https://magoosh.com/gmat/2012/algebra-on ... to-factor/
Let's start with x^2 + ax + 15 = 0. If one root is 3, then one factor must be (x − 3). Let's say the other root is [b]k[/b], and the other factor is (x − k). If we multiply (x - 3)(x - k), the constant term will be (3k), which has to equal 15. Therefore, k = 5, and
(x − 3)(x − 5) = x^2 − 8x + 15
Therefore, a = 8.
Now, we have x^2 + 8x − b = 0, which has equal roots. If a quadratic has equal roots, it must be the square of either a sum or a difference:
Square of a sum: (a + b)^2 = a^2 + 2ab + b^2
Square of a difference: (a − b)^2 = a^2 − 2ab + b^2
See:
https://magoosh.com/gmat/2013/three-alge ... -the-gmat/
Look at the square of a sum formula. We replace a = x. In order for the middle term to equal 8x, we must have b = 4, and the full equation would be
(x + 4)^2 = x^2 + 8x + 16
Setting the final terms equal, we have
− b = 16
b = −16
Answer = (B)
Does all this make sense?
Mike
27 May 2015, 13:52
Kudos
Bookmarks
Bunuel
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.
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.
Hi Bunuel,
I understand how you got to the three roots: 0, 3, and -2 but I missed out on the 0 root because I factored the expression like this:
x^4 = x^3 + 6x^2
x^4 = x^2(x + 6)
x^2 = x + 6
x^2 - x - 6 = 0 .... leading to only roots 3 and -2
I'm not sure why my way is incorrect since. I don't know if I broke some algebra rule. Could you please clarify my factoring method? Thanks, kindly.
