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

Question

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

Bunuel · Accepted Answer

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.

Answer: D.

Bunuel · Answer

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.

varundixitmro2512 · Answer

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

Bunuel · Answer

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KarishmaParmar · Answer

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?

Bunuel · Answer

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 .

sb0541 · Answer

\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 
Answer is D .

Pritishd · Answer

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

Please confirm the OA.

chetan2u · Answer

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

sb0541 · Answer

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 .

ScottTargetTestPrep · Answer

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).

Answer: D

K-ja · Answer

This is one of the best reasons for justifying the elimination of -2.

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

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