Question

Solve the following

(1)x² -|x| - 2 = 0

(2) |x-2| = 1 ​

Answer 1

Hi !

Solution:- (1)

x² -|x|- 2 = 0

=> |x|² - |x| - 2 = 0

=> (|x²| - |x| - 2 = 0

=> |x| = 2

=> x = +-2

Solution :(2)

|x-2| = 1 , i.e those points on real number line which are distance 1 unit from 2 .

<----0-----1-----2----3-->

As shown in the figure x= 1 and x = 3 are distance one units from 2 ; hence

x = 1 or x = 3

thus , |x-2| = 1

or , x -2 = +-1

=> x = 1 or x = 3

__________________________

hope it helps

@Raj❤

Answer 2

(1)x² -|x| - 2 = 0

Let |x| be positive

x > 0

Case (i) :-

x² -|x| - 2 = 0

So, the modulus will open with +ve sign

➡x² - x - 2 = 0

➡x² +x -2x -2 =0

➡x( x +1) -2(x +1) = 0

➡(x-2)(x+1) = 0

Equate the factors to 0 to get the roots of the equation .

x = 2 , -1

Case (ii):-

Let x < 0

x² -|x| - 2 = 0

here, x will open with -ve sign ;

➡x² -(- x) - 2 = 0

➡x² + x - 2 = 0

➡x² -x + 2x -2

➡x(x-1) +2(x-1) = 0

➡(x+2) (x -1) = 0

Equate the factors to 0 to get the roots of the equation .

x = -2 , 1

(2) |x-2| = 1

|x-2| = 1

While, removing modulus ;

x - 2 = ± 1

Case (i) :-

Taking +ve value

x - 2 = + 1

x = 3

Case (ii) :-

Taking -ve value

x - 2 = - 1

x = 1

Answers :- 3 ,1