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.

Hope it helps.
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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

How do you know to break the 39 up into 12, 18, and 9?

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?

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:

\(-2\neq{\sqrt[4]{(-2)^3+6(-2)^2}}\).
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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.
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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...

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?

usre123

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.

Cheers,
Dabral

"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.

Dabral

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 ?

Please explain

Hello, how come m^2n^2 will become (mn)^2 ?

\((mn)^2=(mn)(mn)=m^2*n^2\).
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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.

If you divide (reduce) x^4 = x^2(x + 6) by x^2, you assume, with no ground for it, that x^2 (x) does not equal to zero thus exclude a possible solution.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Hi Bunuel
Could you please check below is fine? Thanks

can we write ? m>0 -----> -x^2-2x+15 > 0
x^2+2x-15 < 0
(x+5)(x-3) < 0
test some big number like 100 --> (x+5)(x-3) will be positive

--- +ve--- -5 ------ -ve------- 3 -----+ve---------

range of x ==> -5<x<3

Hi anupamadw,

Yes, that works. When you have a quadratic inequality and there are two real roots, the number line can be thought of being divided in to three regions. One to the left of x=-5, the region between -5 and 3, and the region to the right of 3. The sign of the expression will change as we go from one region to the next. This means our expression will either follow the path of +, -, + or -, +, - in the three regions. When you chose x=100 as a sample value, we found that the expression (x+5)(x-3) is positive at x=100 and we can conclude that for x>3 it is positive, this means that between -5 and 3 it will be negative, and less -5 it will be positive.

Cheers,
Dabral