IPMAT/Algebra•7mos
solulu

Modulus

2

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Comments (6)

S
scraggy_46
242/360 (IPMAT 2025)
•7mos

Yeah your approach broke because you split it incorrectly.

We need

|x² − 5x + 4| < |x² − 4|

Factor:

x² − 5x + 4 = (x − 1)(x − 4)

x² − 4 = (x − 2)(x + 2)

So inequality becomes

|(x − 1)(x − 4)| < |(x − 2)(x + 2)|

Now square both sides:

(x − 1)²(x − 4)² < (x − 2)²(x + 2)²

Cancel common structure by bringing everything one side and simplifying, you’ll get:

(x − 2)²(x + 2)² − (x − 1)²(x − 4)² > 0

This simplifies to:

(3x − 4)(x − 2)(x − 1) > 0

Now solve sign-wise:

Critical points: x = 1, 4/3, 2

Check intervals → solution comes out as:

x < 1 OR 4/3 < x < 2 OR x > 2

Now restrict to integers in −5 ≤ x ≤ 10:

Valid integers:

−5, −4, −3, −2, −1, 0, 2, 3, 4

Total = 9

1
S
solulu
OP•7mos

I understood this method but why do we have to sqaure the terms?

Is my method wrong?

1
S
solulu
OP•7mos

like over here also they didn't sqaure they just took 2 cases of (x-9) but yes I also know that x-9 has to be considered positive then as it is the output of a modulus

1
S
solulu
OP•7mos

Oh wait I got it. 2 is violating the condition on the left. It will become 0 which isn't allowed.

Thank you!

1
S
solulu
OP•7mos

Yes, but the answer is 9

1
B
bheem0001
•7mos

?you are getting 10 as answer?

1