Question

3.
If the equation x3 - ax2 + bx - a = 0 has three real roots then the
following is true
b=1
a =1
a not equal to 1
b not equal to 1
b=1​

Answer 1

Question: If the equation x³ - ax² + bx - a = 0 has 3 real roots, then which of the following is true.

Options:

• b equal to 1
• a equal to 1
• b not equal to 1
• a not equal to 1

Solution:

Approach: Substitute the values from option in the question. If the solution matches with the conditions in the question, then that is the answer.

Let us substitute b = 1 in the equation since that is the first option.

→ x³ - ax² + x - a = 0

→ x² ( x - a ) + 1 ( x - a ) = 0   [ Factorizing ]

→ ( x² + 1 ) ( x - a ) = 0

For the above set of equations, we get only one real solution 'a'. But according to the question, there must be 3 real solutions to the above equation. Hence b = 1 is not satisfying the equation.

Therefore 'b not equal to 1' is the correct option.

Answer 2

Hence b = 1 does not satisfy the equation x3 - ax2 + bx - a = 0

Step-by-step explanation:

Let us substitute b = 1 in the equation since that is the first option.

→ x³ - ax² + x - a = 0

→ x² ( x - a ) + 1 ( x - a ) = 0   [ Factorizing ]

→ ( x² + 1 ) ( x - a ) = 0

For the given set of equations, we get only one solution i.e. 'a'.

But according to the question, there must be 3 real solutions to the above equation.

Hence b = 1 does not satisfy the equation x3 - ax2 + bx - a = 0