Question

can anyone send soln​

Answer 1

Answer:

1 , 2 , -3

Step-by-step explanation:

First of all, given eq is of 3rd degree.

so, the eq should contain 3 roots.

Let |x| = a

now, the given eq becomes a^3 - 7a + 6 = 0 __(1)

put a = 1 in the above eq(1)

=> 1^3 - 7*1 + 6 = 0 => 1 - 7 + 6 = 0 => 0 = 0

=> a = 1 is one of the roots of eq(1)

By hornor's method,

| 1 0 -7 6

|

1 | 0 1 1 -6

|_______________

| 1 1 -6 0

=> the other two factors lies in a^2 + a - 6 = 0

=> a^2 + 3a - 2a - 6 = 0

=> a ( a + 3 ) - 2 ( a + 3 ) = 0

=> ( a + 3 ) ( a - 2 ) = 0

=> a + 3 = 0 or a - 2 = 0

=> a = -3 or a = 2

but |x| = a => x = +a or -a

the values which satisfies the given eq are 1,2&-3

.•. the solution is x = 1 , 2 , -3