Sum of Roots

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by theCodeToGMAT » Tue Oct 29, 2013 10:56 pm
|x + 4|^2 - 10|x + 4| = 24

Let |x+4| = A

A^2 -10A - 24=0

A = 12 & -2

|x+4| = 12

x+4 = 12 ==> 8
Or, -x - 4 = 12 ==> -16

Sum = 8-16 = -8

Here, we ignored A = -2 because A cannot be negative.
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by ganeshrkamath » Wed Oct 30, 2013 2:57 am
[email protected] wrote:What is the sum of all roots of the equation
|x + 4|^2 - 10|x + 4| = 24?
|x + 4|^2 - 10|x + 4| = 24

Let |x+4| = P

P^2 - 10P - 24 = 0
Roots of equation of the form ax^2 + bx + c = 0 are (-b +/- sqrt(b^2 - 4ac))/2a

Here a = 1, b = -10. and c = -24

So the roots are
(10 + sqrt(100 + 2*24))/2 and (10 - sqrt(100 + 2*24))/2
= (10 + sqrt(100 + 96))/2 and (10 - sqrt(100 + 96))/2
= (10 + 14)/2 and (10 - 14)/2
= 24/2 and -4/2
= 12 and -2

But, P = |x+4| >= 0

|x+4| = 12

(x+4) = 12 or (x+4) = -12
x = 8 or x = -16

Sum of the roots = 8 - 16 = -8

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by GMATGuruNY » Wed Oct 30, 2013 3:19 am
[email protected] wrote:What is the sum of all roots of the equation
|x + 4|^2 - 10|x + 4| = 24?
Alternate approach:

|x+4| represents a NONNEGATIVE value.

Case 1: x≥-4
Here, |x+4| = x+4.
To illustrate:
If x=-3, then |x+4| = x+4 = -3+4 = 1.
The result is a nonnegative value.

Case 2: x<-4
Here, |x+4| = -x-4.
To illustrate:
If x=-5, then |x+4| = -x-4 = -(-5)-4 = 1.
The result is a nonnegative value.

Solve for each case.

Case 1: x≥-4, implying that |x+4| = x+4
(x+4)² - 10(x+4) = 24
x² + 8x + 16 - 10x - 40 - 24 = 0
x² - 2x - 48 = 0
(x+6)(x-8) = 0.
x=-6 or x=8.
But x=-6 is not valid, since the range here is x≥-4.
Thus, the only valid solution in Case 1 is x=8.

Case 2: x<-4, implying that |x+4| = -x-4
(-x-4)² - 10(-x-4) = 24
x² + 8x + 16 + 10x + 40 - 24 = 0
x² + 18x + 32 = 0
(x+2)(x+16) = 0
x=-2 or x=-16.
But x=-2 is not valid, since the range here is x<-4.
Thus, the only valid solution in Case 2 is x=-16.

Sum of the solutions = 8 + (-16) = -8.

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by Matt@VeritasPrep » Wed Oct 30, 2013 10:22 pm
[email protected] wrote:What is the sum of all roots of the equation
|x + 4|^2 - 10|x + 4| = 24?
Factoring is probably the way, and you don't have to change your variables as was done in some earlier explanations. One neat thing about a question such as this is that you can factor it as you would a normal quadratic: just treat |x+4| like the customary x. For example, if you had

x² - 10x - 24 = 0

you would factor that as

(x - 12)(x + 2) = 0

So here we have

|x+4|² - 10|x+4| - 24 = 0

or

(|x+4| - 12)(|x+4| + 2) = 0

So either |x+4| = 12 or |x+4| = -2.

The second one is impossible, so just solve the first one |x+4| = 12 and you're done!