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BrentGMATPrepNow
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x Updated on: 19 Feb 2025, 12:42
Originally posted by BrentGMATPrepNow on 02 Jan 2017, 13:45.
Last edited by Bunuel on 19 Feb 2025, 12:42, edited 3 times in total.
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95% (hard)

Question Stats:

27% (02:04) correct 73% (02:06) wrong based on 422 sessions

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What is the sum of all solutions to the equation \(x^{(2x^2 + 4x – 6)} = x^{(x^2 + 8x +6)}\) ?

A) -4
B) -3
C) 3
D) 4
E) 5
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E
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BrentGMATPrepNow
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 03 Jan 2017, 11:20
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ -1
For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. For example, 1² = 1³, but we can't conclude that 2 = 3.

So, let's first see what happens when the base (x) equals 0, 1 and -1

If x = 0, then we have: 0^(2(0²) + 4(0) – 6) = 0^(0² + 8(0) + 6)
Simplify: 0^(-6) = 0^6
Evaluate: 0 = 0
So, x = 0 is one solution to the equation (yes, I know that x = 0 does not change the SUM of the solutions. I just want to show all of the possible considerations)

If x = 1, then we have: 1^(2(1²) + 4(1) – 6) = 1^(1² + 8(1) + 6)
Simplify: 1^0 = 1^15
Evaluate: 1 = 1
So, x = 1 is another solution to the equation

If x = -1, then we have: (-1)^[2(-1)² + 4(-1) – 6] = (-1)^[(-1)² + 8(-1) + 6]
Simplify: (-1)^(-8) = (-1)^(-1)
Evaluate: 1 = -1
So, x = -1 is NOT a solution to the equation

Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation.
Given: x^(2x² + 4x – 6) = x^(x² + 8x + 6)
Since the bases are the same, we can write: 2x² + 4x – 6 = x² + 8x + 6
Rearrange to get: x² - 4x – 12 = 0
Factor to get: (x - 6)(x + 2) = 0
So, x = 6 and x = -2 are also solutions to the equation.

So, the solutions are x = 0, x = 1, x = 6, and x = -2
0 + 1 + 6 + (-2) = 5

Answer: E
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 02 Jan 2017, 16:49
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 02 Jan 2017, 16:58
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D

That's close, but you haven't found all of the possible solutions.

Cheers,
Brent
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 02 Jan 2017, 18:41
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D

That's close, but you haven't found all of the possible solutions.

Cheers,
Brent

I see it now, should have noted that 1^0=1^15.
Add 1 to 4 and the answer is 5.
Thanks. Good problem.
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 02 Jan 2017, 20:19
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gracie
I see it now, should have noted that 1^0=1^15.
Add 1 to 4 and the answer is 5.
Thanks. Good problem.

Nice work.
I should also note that x = 0 is another solution (but it doesn't change the answer)

We must also check to see whether x = -1 is a possible solution.
When we plug x = -1 into the equation, we find that it is NOT a solution.

So, the solutions are x = 6, x = -2 and x = 1

Cheers,
Brent
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 20 Mar 2018, 10:49
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

The equation can be written as -

\(x ^{(x-6)(x+2)} = 1\)

L.H.S = R.H.S, when x = 1,6,-2
Hence, Sum = 5
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 24 Mar 2018, 02:24
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Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 08 May 2018, 13:11
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Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0

Unfortunately, you won't have a calculator (or a log table for that day) on test day.
That said, if you did have a calculator, what values would you plug in for x?

Cheers,
Brent
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 08 May 2018, 13:24
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srinjoy1990
Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0

Unfortunately, you won't have a calculator (or a log table for that day) on test day.
That said, if you did have a calculator, what values would you plug in for x?

Cheers,
Brent


we need not take values thats the point, we can solve the equation by taking log.

you have something like this,

logx(x^2-4x-12) = 0.

logx =0 | x^2-4x-12=0

so, roots are, x=1 , and the solution to the equation x^2-4x-12=0.

so, 5 (1+4).
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 26 Jun 2018, 05:09
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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions
x^(2x² + 4x – 6) = x^(x² + 8x +6)
2x² + 4x – 6 = x² + 8x +6
x² - 4x - 12 = (x+2)(x-6)
from here x = -2 , 6
here x = 1 will also satisfy the equation
so -2+6+1 = 5
E is the answer
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 06 Nov 2019, 10:59
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x = 1,6,-2
Sum = 5.
Hence E.

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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 11 Apr 2021, 06:08
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\(x^{(2x^2 + 4x – 6)} = x^{(x^2 + 8x +6)}\)

Bring both equations on same side and this becomes -

\(x^{(2x^2 + 4x – 6) - (x^2 + 8x +6)}\)

\(x^{(x^2 - 4x – 12)}\) = 1

Now any number to power another number is equal to 1 in 2 cases
\(n^{(0)}\) or n = 1

So, lets check both cases
Case 1 -
\(n^{(0)}\) which means => \(x^2 - 4x – 12 = 0\)
(x-6)(x+2) = 0 i.e. x = 6, -2

Case 2 -
x = 1

So sum of all -> 6-2+1 = 5
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 19 Feb 2025, 12:39
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I dont understand 1 please elaborate
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x Updated on: 19 Feb 2025, 21:53
Originally posted by Kinshook on 19 Feb 2025, 21:50.
Last edited by Kinshook on 19 Feb 2025, 21:53, edited 1 time in total.
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What is the sum of all solutions to the equation \(x^{(2x^2 + 4x – 6)} = x^{(x^2 + 8x +6)}\) ?

x = 0; LHS= 0ˆ{-6} is invalid
x = 1; 1ˆ0 = 1ˆ15 = 1; TRUE
x = -1; LHS = (-1)ˆ{-8} = 1; RHS = (-1)ˆ{-1} = -1; LHS <> RHS; NOT TRUE

Other solutions: -
\(2x^2 + 4x - 6 = x^2 + 8x + 6\)
\(x^2 - 4x - 12 = 0\)
(x-6)(x+2) = 0
x = 6 or
x = - 2

x = {-2,1,6}
Sum of all solutions to the equation = -2 + 1+ 6 = 5

IMO E
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What is the sum of all solutions to the equation x^(2x2 + 4x 6) = x 19 Feb 2025, 21:53
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BrentGMATPrepNow

\(0^{-6} = \frac{1}{0^6} = \frac{1}{0}\) is undefined.
x = 0 is NOT a solution

Please correct
BrentGMATPrepNow
GMATPrepNow
What is the sum of all solutions to the equation x^(2x2 + 4x – 6) = x^(x2 + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ -1
For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. For example, 12 = 13, but we can't conclude that 2 = 3.

So, let's first see what happens when the base (x) equals 0, 1 and -1

If x = 0, then we have: 0^(2(02) + 4(0) – 6) = 0^(02 + 8(0) + 6)
Simplify: 0^(-6) = 0^6
Evaluate: 0 = 0
So, x = 0 is one solution to the equation (yes, I know that x = 0 does not change the SUM of the solutions. I just want to show all of the possible considerations)

If x = 1, then we have: 1^(2(12) + 4(1) – 6) = 1^(12 + 8(1) + 6)
Simplify: 1^0 = 1^15
Evaluate: 1 = 1
So, x = 1 is another solution to the equation

If x = -1, then we have: (-1)^[2(-1)2 + 4(-1) – 6] = (-1)^[(-1)2 + 8(-1) + 6]
Simplify: (-1)^(-8) = (-1)^(-1)
Evaluate: 1 = -1
So, x = -1 is NOT a solution to the equation

Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation.
Given: x^(2x2 + 4x – 6) = x^(x2 + 8x + 6)
Since the bases are the same, we can write: 2x2 + 4x – 6 = x2 + 8x + 6
Rearrange to get: x2 - 4x – 12 = 0
Factor to get: (x - 6)(x + 2) = 0
So, x = 6 and x = -2 are also solutions to the equation.

So, the solutions are x = 0, x = 1, x = 6, and x = -2
0 + 1 + 6 + (-2) = 5

Answer: E
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