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What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 36% (01:46) correct 64% (01:31) wrong based on 382 sessions

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What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

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Originally posted by GGMU on 11 Jul 2016, 01:37.
Last edited by Bunuel on 11 Jul 2016, 01:45, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 59020
Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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4
7
anurag16 wrote:
What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

First of all notice that since LHS is the square root of a number, it must be non-negative (the square root function cannot give negative result), then the RHS must also be non-negative: $$x+1\geq 0$$ --> $$x \geq -1$$.

Square the equation: $$2x^2-x-9=x^2+2x+1$$ --> $$x^2-3x-10=0$$ --> x=-2 or x=5. Discard x=-2 because it's not >=-1. We are left with only one root: 5.

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Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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Hi Bunuel,

please explain why cant we have a neg number after rooting a number

root of 16 can be -4 or +4. Thats what we also apply in different DS quest
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Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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Bunuel wrote:
anurag16 wrote:
What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

First of all notice that since LHS is the square root of a number, it must be non-negative (the square root function cannot give negative result), then the RHS must also be non-negative: $$x+1\geq 0$$ --> $$x \geq -1$$.

Square the equation: $$2x^2-x-9=x^2+2x+1$$ --> $$x^2-3x-10=0$$ --> x=-2 or x=5. Discard x=-2 because it's not >=-1. We are left with only one root: 5.

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Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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varundixitmro2512 wrote:
Hi Bunuel,

please explain why cant we have a neg number after rooting a number

root of 16 can be -4 or +4. Thats what we also apply in different DS quest

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{16}=4$$, NOT +4 or -4. In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4. Even roots have only a positive value on the GMAT.
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GMAT 1: 710 Q48 V40 Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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1
Bunuel wrote:
anurag16 wrote:
What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

First of all notice that since LHS is the square root of a number, it must be non-negative (the square root function cannot give negative result), then the RHS must also be non-negative: $$x+1\geq 0$$ --> $$x \geq -1$$.

Square the equation: $$2x^2-x-9=x^2+2x+1$$ --> $$x^2-3x-10=0$$ --> x=-2 or x=5. Discard x=-2 because it's not >=-1. We are left with only one root: 5.

Hello , I have a doubt related to this question. Is it only the even roots that do not yield negative numbers or any number that's under root of another number can't be negative on the GMAT?
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Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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KarishmaParmar wrote:
Bunuel wrote:
anurag16 wrote:
What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

First of all notice that since LHS is the square root of a number, it must be non-negative (the square root function cannot give negative result), then the RHS must also be non-negative: $$x+1\geq 0$$ --> $$x \geq -1$$.

Square the equation: $$2x^2-x-9=x^2+2x+1$$ --> $$x^2-3x-10=0$$ --> x=-2 or x=5. Discard x=-2 because it's not >=-1. We are left with only one root: 5.

Hello , I have a doubt related to this question. Is it only the even roots that do not yield negative numbers or any number that's under root of another number can't be negative on the GMAT?

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{16}=4$$, NOT +4 or -4. In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4. Even roots have only a positive value on the GMAT.

Odd roots have the same sign as the base of the root. For example, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.
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Re: Very hard - quant problem  [#permalink]

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1
1
nive28 wrote:
What is the sum of all possible solutions to the equation \sqrt{2x^2-x-9}=x+1?

1. -2
2. 2
3. 3
4. 5
5. 6

$$\sqrt{2x^2-x-9}$$ = x+1
squaring both sides
$$x^2-3x-10=0$$
(x-5)(x+2)=0
x = 5 or -2
sum of the possible solutions is 5 + (-2) = 3
But -2 is not a solution as it will not satisfy the equation . so , only 5 is valid solution

Originally posted by sb0541 on 01 Jan 2017, 06:39.
Last edited by sb0541 on 01 Jan 2017, 12:37, edited 1 time in total.
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Re: Very hard - quant problem  [#permalink]

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I am assuming this question to be $$\sqrt{2x^2-x-9}=x+1$$

Squaring on both the sides:
$$2x^2-x-9=x^2+2x+1$$
Simplifying will yield:
$$x^2-3x-10=0$$
$$(x-5)(x+2)$$
x = 5 or -2
Sum of all possible solutions is $$5 + (-2) = 3$$
Ans: C

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Originally posted by Pritishd on 01 Jan 2017, 06:41.
Last edited by Pritishd on 01 Jan 2017, 06:42, edited 1 time in total.
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Posts: 8157
Re: Very hard - quant problem  [#permalink]

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nive28 wrote:
What is the sum of all possible solutions to the equation \sqrt{2x^2-x-9}=x+1?

1. -2
2. 2
3. 3
4. 5
5. 6

One way to do it..
$$\sqrt{2x^2-x-9}=x+1$$...
Square both sides..
$$2x^2-x-9=x^2+2x+1........ x^2-3x-10=0$$
This is of the form ax^2+bx+c=0...
sum of roots: -b/a
Product of roots: C/a

Here b is -3 and a is 1..
Ans =-(-3)/1=3

On second thought
However the answer may not come correctly here.
$$x^2-3x-10=0.......(x-5)(x+2)=0$$
So the solutions are 5 and -2...
But in initial equation, x as -2 will give x+1as -2+1=-1..
But in GMAT square root is always POSITIVE, so -2 cannot be a solution..
Only 5 is possible..
D

Good Q
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Re: Very hard - quant problem  [#permalink]

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nive28 wrote:
Pritishd wrote:
I am assuming this question to be $$\sqrt{2x^2-x-9}=x+1$$

Squaring on both the sides:
$$2x^2-x-9=x^2+2x+1$$
Simplifying will yield:
$$x^2-3x-10=0$$
$$(x-5)(x+2)$$
x = 5 or -2
Sum of all possible solutions is $$5 + (-2) = 3$$
Ans: C

OA: option d: 5

If we put back the solution x = -2 into the equation then it is not valid .
We get 1 = -(1) , which is not true .
However, for x = 5 then equation is valid giving the result 6 = 6 .

Hence only 5 is a valid solution for the equation .
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Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^  [#permalink]

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anurag16 wrote:
What is the sum of all possible solutions to the equation $$\sqrt{2x^2-x-9}=x+1$$?

a. -2
b. 2
c. 3
d. 5
e. 6

Squaring the equation we have:

2x^2 - x - 9 = (x + 1)^2

2x^2 - x - 9 = x^2 + 2x + 1

x^2 - 3x - 10 = 0

(x - 5)(x + 2) = 0

x = 5 or x = -2

However, whenever we are dealing with an equation involving square root(s), we always have to check our solutions.

If x = 5, we have:

√(2(5)^2 - 5 - 9) = 5 + 1 ?

√(36) = 6 ?

6 = 6 ? Yes!

If x = -2, we have:

√(2(-2)^2 - (-2) - 9) = -2 + 1 ?

√(1) = -1 ?

1 = -1 ? No!

Thus we see that the only solution is 5 (and thus the sum of all possible solutions is 5).

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