Question

The real roots of | x |³ - 3 x² + 3 | x | - 2 = 0 are
(a) 0, 2 (b) ±1 (c) ±2 (d) 1, 2

Answer 1

Question:

The real roots of |x|³ - 3x² + 3|x| - 2 = 0 are :

a) 0 , 2

b) ±1

c) ±2

d) 1,2

Answer:

c) ± 2

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

• If |x| = x ig x ≥ 0 and |x| = -x if x < 0 .

• |x| = max{ x , -x }

• If |x| = a (where a ≥ 0) , then x = ± a .

• x² = |x|²

Solution:

Here,

The given equation is : |x|³ - 3x² + 3|x| - 2 = 0

We know that , x² = |x|²

Thus,

The given equation may be written as :

|x|³ - 3|x|² + 3|x| - 2 = 0 ---------(1)

Now,

Let |x| = y , then the eq-(1) will reduce to ;

y³ - 3y² + 3y - 2 = 0 --------(2)

Now,

Let's find the roots of eq-(2) by splitting the terms and factorising it further.

Thus,

=> y³ - 3y² + 3y - 2 = 0

=> y³ - 2y² - y² + 2y + y - 2 = 0

=> y²(y-2) - y(y-2) + (y-2) = 0

=> (y-2)(y²-y+1) = 0

Case1 : y - 2 = 0

=> y - 2 = 0

=> y = 2

=> |x| = 2

=> x = ± 2

Case2 : y² - y + 1 = 0

( This is not possible )

The discriminant, D = (-1)² - 4•1•1 = -3 which is less then zero.

Since , D < 0 , thus there doesn't exist any real value of y for which y² - y + 1 = 0.

Hence,

There are two distinct roots of the given equation, namely x = ± 2 .