If 25|x^2−5x−10|=100, then what is the sum of all the possible values

It is 10.

|expression| = 4
exp = +4 => on solving x=7 and -2
exp = -4 => on solving x = 3 and 2
7 - 2 + 2 + 3 = 10

updated *roots for 2nd exp are wrong, that should be 6 or -1 instead of +3 or -2*

7-2+6-1 = 10 is correct


oa pls

\(25|x^2−5x−10|=100\)

|x^2−5x−10|=4

x^2−5x−10=4
x=7 & -2
or say
-(x^2−5x−10)=4
x=3,+2

sum = 7-2+2+3 = 10

IMO A

Solution

Given:

• \(25|x^2 - 5x - 10| = 100\)

To find:

• The sum of all the possible values of x

Approach and Working:
\(|x^2 - 5x - 10| = \frac{100}{25}= 4\)

• This implies, \(x^2 - 5x – 10 = 4\) or \(x^2 - 5x – 10 = -4\)

o \(x^2 - 5x – 14 = 0\), implies, x = 7 or -2
o \(x^2 - 5x – 6 = 0\), implies, x = 6 or -1

Therefore, sum of all possible values of x = 7 – 2 + 6 – 1 = 10

Hence, the correct answer is Option A

Answer: A

Just solve both the quadratic expressions formed when we open the modulus

\(x^2 - 5x -14 = 0\), will give x as -2 and 7

\(x^2 - 5x + 6 = 0\), will give x as 2 and 3

Add all the values of x to get 10

A

for the equation ax^2 + bx + c = 0 ,the sum of roots is given by (-b/a)

now acc. to ques:
|x^2−5x−10|=4

=> x^2−5x−10= 4 ...... eq(1)
and x^2−5x−10= -4 ...... eq(2)

now from eq (1) and eq (2)

r1 + r2 = -(-5/1)= 5
r3 + r4 = -(-5/1)= 5

therefore , r1 + r2 + r3 + r4= 10

\(25|x^2−5x−10|=100\)

i.e. \(|x^2−5x−10|=4\)

i.e. \(x^2−5x−10=4\) or i.e. \(x^2−5x−10=-4\)

i.e. \(x^2−5x−14=0\) or i.e. \(x^2−5x−6=0\)

i.e. \(x^2−7x+2x−14=0\) or i.e. \(x^2−6x+x−6=0\)

i.e. \((x+2)*(x-7)=0\) or i.e. \((x−6)*(x+1)=0\)

i.e. x = -2, 7, 6, -1

Aum of all values of x = 10

Answer: Option A

hello folks!

if we apply the 'open up modulus' method, and find the solutions from both equations (i.e. 6, -1, 7, -2), do we need to test each one solution in the formula and check if they're correct?

or can we just take the solutions and sum them up without checking?

thanks!

x^2-5x-10=4... Sum is -b/a=5
x^2-5x-10=-4... Sum=5
So sum of all possible values is 10

Posted from my mobile device

x^2-5x-10=4... Sum is -b/a=5
x^2-5x-10=-4... Sum=5
So sum of all possible values is 10

Posted from my mobile device

|x^2−5x−10|=4

x^2−5x−10=4
x=7 & -2
or say
-(x^2−5x−10)=4
x=3,+2

sum = 7-2+2+3 = 10

correct answer A

Hi All,

I've been using TTP for my quant prep and I'm having some difficulties in understanding this concept. Apparently when solving for for equations with absolute values, I need to split the equation into positive and negative then find all of the potential solutions. Then, I need to plug the solutions back into the original equation to determine what the actual solutions are. However, in the practice question solutions, TTP only does the check half of the time.

For this question specifically, I keep trying to plug the potential solutions into the original equation and I keep getting 3 of the potential solutions as false. However, that is not the answer provided by TTP. Could someone help me solve this?

If 25 |x^2 - 5x - 10| = 100, then what is the sum of all the possible values of X?

A. 10
B. 8
C. 2
D. -4
E. -8



Thank you so much,

25 |x^2 - 5x - 10| = 100 => |x^2-5x-10| = 4

We have 2 cases:

Case 1: x^2-5x-10 >0
=> x^2-5x-10 = 4
=> x^2-5x-14 =0
=>x^2 - 7x+2x-14 =0
=>x(x-7) +2(x-7) =0
=>(x+2)(x-7)=0
=>x=-2 or x=7

Case 2: x^2-5x-10 <0
=> x^2-5x-10 = -4
=> x^2-5x-6 =0
=>x^2-6x+x-6=0
=>x(x-6)+1(x-6)=0
=>(x+1)(x-60=0
=>x=-1 or x=6

Adding all 4 values -2+7+6-1 = 10
Hence A


For plugging in values:

25 |x^2 - 5x - 10| = 100 =>

|x^2-5x-10| = 4

x=7, |49-35-10| = |49-45| = 4 Correct
x= 6, |36 -30 -10| =|-4| =4 Correct
and x = -1, |1+5-10| = |-4| = 4 Correct

We can rewrite the equations as |x^2 - 5x - 10| = 4

x^2 - 5x - 10 = 4
(or)
x^2 - 5x - 10 = -4

x^2 - 5x - 14 = 0

or

x^2 - 5x - 6 = 0

solving each equations separately,

(x-7)(x+2) = 0

or

(x-6)(x+1) = 0

so possible values of x is 7, -2, 6, -1.

Sum of all possible values = 7-2+6-1 = 10.

So the correct choice is A

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