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

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

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:
https://magoosh.com/gmat/2012/gmat-quant-roots/

Does all this make sense?
Mike

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

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.

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