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If x-5 = (2x^2-18x+37)^1/2 then x could equal

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If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post Updated on: 05 Dec 2013, 02:49
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Question Stats:

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If \(x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

A. 2
B. 3
C. 4
D. 5
E. 6

Now we get 2 & 6. If we put 2 in the equation we get \(3=\sqrt{9}\)

so shouldn't the answer be 2 and not 6?

Originally posted by theGame001 on 04 Dec 2013, 14:19.
Last edited by Bunuel on 05 Dec 2013, 02:49, edited 1 time in total.
Renamed the topic and edited the question.
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Re: if x-5 = [#permalink]

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New post 04 Dec 2013, 14:52
2
theGame001 wrote:
\(If x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

a) 2
b) 3
c) 4
d) 5
e) 6

Now we get 2 & 6. If we put 2 in the equation we get \(3=\sqrt{9}\)

so shouldn't the answer be 2 and not 6?

Dear theGame001,
I'm happy to help. :-)

You may find this blog helpful:
http://magoosh.com/gmat/2013/gmat-math- ... -radicals/

This might just be a small oversight, but if we put in x = 2, the left side becomes
2 - 5 = -3
The left side is (-3) and the right side is sqrt(9) = +3, so they are not equal.

If we put in x = 6, then on the left side we get 6 - 5 = +1.
On the right side, sqrt(1) = +1
so, they are equal.

Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: if x-5 = [#permalink]

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New post 04 Dec 2013, 14:56
mikemcgarry wrote:
theGame001 wrote:
\(If x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

a) 2
b) 3
c) 4
d) 5
e) 6

Now we get 2 & 6. If we put 2 in the equation we get \(3=\sqrt{9}\)

so shouldn't the answer be 2 and not 6?

Dear theGame001,
I'm happy to help. :-)

You may find this blog helpful:
http://magoosh.com/gmat/2013/gmat-math- ... -radicals/

This might just be a small oversight, but if we put in x = 2, the left side becomes
2 - 5 = -3
The left side is (-3) and the right side is sqrt(9) = +3, so they are not equal.

If we put in x = 6, then on the left side we get 6 - 5 = +1.
On the right side, sqrt(1) = +1
so, they are equal.

Does all this make sense?
Mike :-)


yes it definitely does. My bad. Thank you very much.
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Re: If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post 05 Dec 2013, 23:35
2
1
theGame001 wrote:
If \(x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

A. 2
B. 3
C. 4
D. 5
E. 6

Now we get 2 & 6. If we put 2 in the equation we get \(3=\sqrt{9}\)

so shouldn't the answer be 2 and not 6?



We know that the term \sqrt{2x^2-18x+37} is greater than or equal to 0--------> this implies LHS is greater than or equal to zero and hence Option A,B and C are ruled out as LHS becomes negative for any of the above 3 values.

If you take x =5 then LHS is 0 but RHS is \sqrt{3}. If we put x= 6 then we get LHS=RHS=1

Ans E
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Re: If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post 17 Jun 2015, 12:18
mikemcgarry wrote:
theGame001 wrote:
\(If x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

a) 2
b) 3
c) 4
d) 5
e) 6

Now we get 2 & 6. If we put 2 in the equation we get \(3=\sqrt{9}\)

so shouldn't the answer be 2 and not 6?

Dear theGame001,
I'm happy to help. :-)

You may find this blog helpful:
http://magoosh.com/gmat/2013/gmat-math- ... -radicals/

This might just be a small oversight, but if we put in x = 2, the left side becomes
2 - 5 = -3
The left side is (-3) and the right side is sqrt(9) = +3, so they are not equal.

If we put in x = 6, then on the left side we get 6 - 5 = +1.
On the right side, sqrt(1) = +1
so, they are equal.

Does all this make sense?
Mike :-)


But if we decide to square both sides:

LHS we get / RHS we get
\(x^2-10x+25 = 2x^2-18x+37\)
\(x^2-10x=2x^2-18x+12\)
\(-10x=x^2-18x+12\)
\(0=x^2-8x+12\)

If you plug in 2, then the equation stays true???
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Re: If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post 18 Jun 2015, 10:19
1
reto wrote:
But if we decide to square both sides:

LHS we get / RHS we get
\(x^2-10x+25 = 2x^2-18x+37\)
\(x^2-10x=2x^2-18x+12\)
\(-10x=x^2-18x+12\)
\(0=x^2-8x+12\)

If you plug in 2, then the equation stays true???

Dear reto,
I'm happy to respond. :-)

My friend, ordinarily when we do the same thing to both sides of an algebraic equation, the fundamental mathematical relationships remain unchanged, and the equation would retain the same roots. A big exception is when we square both sides of an equation with a radical: this situation has the potential to create "extraneous roots," that is, roots that work in the new squared equation but not in the original radical equation. The value x = 2 is just such an extraneous root. It is absolutely a solution for the squared equation, but it does not work in the original equation. It is a solution that has been artificially added by the act of squaring both sides.

Think of a much simpler example. Suppose we have the ridiculously easy equation:
\(sqrt(25) = x\)
Obviously, the only solution to this is x = 5. Now, suppose we square both sides. We get:
\(x^2 = 25\)
and this new, squared equation has two roots, x = +5 and x = -5. The latter is an extraneous root that does not hold in the original equation but emerges as an extra solution when we square both sides.

For more, see:
http://magoosh.com/gmat/2012/gmat-quant-roots/

Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post 17 Sep 2017, 17:03
theGame001 wrote:
If \(x - 5 = \sqrt{2x^2-18x+37}\) then x could equal

A. 2
B. 3
C. 4
D. 5
E. 6

If you don't catch WoundedTiger 's solution (by far the fastest), and you do the algebra . . .

Square both sides. Check solutions. RHS and LHS cannot be negative. RHS has "the square root." By convention, "the square root" refers to the positive square root ONLY.

\(x - 5 = \sqrt{2x^2-18x+37}\)

\((x - 5)^{2} = (\sqrt{2x^2-18x+37})^{2}\)

\(x^2 - 10x + 25 = 2x^2 - 18x + 37\)

\(x^2 - 8x + 12 = 0\)

\((x - 6)(x - 2) = 0\)

\(x = 6\) or \(x = 2\)

LHS and RHS must be nonnegative or positive.

RHS cannot be negative. The square root sign denotes only the positive square root.

If x = 2, LHS is negative. (2 - 5) = -3.

Not possible. LHS must = RHS, and RHS cannot be negative.

Or \(\sqrt{x^{2}} = |x|\), such that RHS is nonnegative or positive, and LHS must be, too, in order for sides to be equal. LHS cannot be negative.

x = 6

ANSWER E
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Re: If x-5 = (2x^2-18x+37)^1/2 then x could equal [#permalink]

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New post 18 Sep 2017, 20:18
One more option is to plug in answer choices as the value of x in the above equation.
First Hint is that RHS cannot be -ve as its a square root.
The only option left is 6
6-5=√2(6)²-18(6)+37
1=√1
1=1(√1=1)
we actually need not do this calculation.
Re: If x-5 = (2x^2-18x+37)^1/2 then x could equal   [#permalink] 18 Sep 2017, 20:18
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