Question

(-p²) - 4.3.2 > 0 real roots ​

Answer 1

Step-by-step explanation:

Given equation is ∣x−3∣

2

+∣x−3∣−2=0,

To solve this equation consider ∣x−3∣=a,

a

2

+a−2=0

⟹(a+2)(a−1)=0

⟹a=−2 or a=1,

But ∣x−3∣ cannot be a negitive number,

⟹∣x−3∣=1

⟹x−3=±1

⟹x=4,2,

∴ sum of all the real roots is 4+2=6