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The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

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.

x cannot be negative cannot be negative as it equals to the even (4th) root of some expression.
_________________

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 \(\sqrt[3]{-64} =-4\).

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 \(x=3\).

Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz

If the entire term under the 4th sq root , say is 16 then my x can be a -2 ryt?
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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 \(x=3\).

Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz

If the entire term under the 4th sq root , say is 16 then my x can be a -2 ryt?

\(\sqrt[4]{16}=2\), not +2 or -2.

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 non-negative value on the GMAT.

Please explain .. not able to get it through !! [#permalink]

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03 Aug 2014, 07:11

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

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, This is a tricky question, and I am happy to help.

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: http://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
_________________

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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).

Answer: B.

Is there another solution in which we don't need to manipulate the number so much? How do you know when there is a shortcut like used here and when there isn't?

[ 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, Request you to please let me know where I'm making a mistake.

Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\)

i.e \((x-0)^2(x-(-6))>=0\) This implies that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) Please let me know where I am going wrong.

Thanks

Plug x=-2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?

well u can plug in -2 and still it holds.

x^4 = (-2)^4 = 16 on the other side of equal to ie x^3+6x^2 we can write (-2)^3+6(-2)^2 equals -8+24 = 16 .....so holds ....please correct me . Also BTW in solving quadratic equation such as x^2-x+6=0 we do consider both the roots -ve and +ve , until its stated not to ... thanks

Hi Bunuel, Request you to please let me know where I'm making a mistake.

Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\)

i.e \((x-0)^2(x-(-6))>=0\) This implies that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) Please let me know where I am going wrong.

Thanks

Plug x=-2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?

well u can plug in -2 and still it holds.

x^4 = (-2)^4 = 16 on the other side of equal to ie x^3+6x^2 we can write (-2)^3+6(-2)^2 equals -8+24 = 16 .....so holds ....please correct me . Also BTW in solving quadratic equation such as x^2-x+6=0 we do consider both the roots -ve and +ve , until its stated not to ... thanks

No, it does NOT hold. You should plug into the original equation as I mention in my post you quote:

These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.

These questions are not from GMATPrep Exam Pack 1. I know #6 is an exact copy of an official GMAT questions, but the others are not official GMAT questions.

Dabral

Those are GMAT Club Tests question. Changed the tag. Thanks.
_________________

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.

Hello, Can you pls tell me my error? I took \(x^2\) common and got \(x^2(x+6)\) = \(x^4\) then i divided everything by \(x^2\). So I only get two roots. x=-2 and x=3. I don't get x=0...

By dividing both sides by \(x^2\) you have eliminated the solution \(x=0\).

For example, if we are given the condition \(x(x-1)=0\), then there are two values of \(x\) that satisfies this equation, \(x=0\) and \(x=1\). However, if we divide both sides by \(x\) we end up with \(x-1=0\), and left with only one solution. In general, on the GMAT never divide by a variable, unless you know that it is non-zero.

"Again, I did something basic wrong. I left 12 on the RHS, and took mn common to have mn(mn+1)=12. mn=12 or mn=11....What is wrong with my concept? "

You will find the mn=12 or mn=11 do not satisfy the above equation. We are looking for the same value of mn, this one is easy enough that many people can guess that mn=3 works, because (3)(3+1)=12, and the other solution is mn=4, which is a bit harder to spot. The better approach is to write it as a quadratic and then factor as shown in the solution.

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, It is understood that even root of a negative expression is not possible. But here, for x= -2, the expression under root, x^3+6x^2 = -8 + 24 = 16 which is positive. Then why x = -2 invalid ?