Question
The number of real solutions of the equation |x|2−3|x|+2=0 are
Answer: Option D
:
D
Given|x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: x < 0 This gives x2+3x+2=0
⇒ (x+2)(x+1)=0⇒ x=−2,−1
Also x=−1,−2 satisfy x<0, so x=−1,−2 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒x =±1, x =±2.
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:
D
Given|x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: x < 0 This gives x2+3x+2=0
⇒ (x+2)(x+1)=0⇒ x=−2,−1
Also x=−1,−2 satisfy x<0, so x=−1,−2 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒x =±1, x =±2.
Was this answer helpful ?
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