Question

The equation x6-x-1=0 has, (a) No positive real roots. (b) Exactly one positive real root. (c) Exactly two positive real roots (d) All positive real roots. Answer OA В с D Submit​

Answer 1

Answer:

x = 0, x= 2/5

Step-by-step explanation:

x6 - x - 1 = 0

x( x5 -1 -1) = 0

x( x5 - 2) = 0

x has two values

Answer 2

Given:

x6-x-1=0

To find:

The nature of the roots of the equation

Solution:

The equation x6-x-1=0 has exactly two positive real roots. (Option c)

We can find the nature by following the given steps-

We know that the nature of the roots of an equation can be determined by finding the value of the determinant of the equation.

The determinant of an equation, D= b²-4ac

Here a is the coefficient of x², b is the coefficient of x, and c is the constant in the equation.

The equation:

${x}^{6} - x - 1 = 0$

It can also be written as (x³)²-x-1=0

So, the value of a=1, the value of b= -1, and the value of c= -1.

Now we will calculate the value of the determinant.

D=b²-4a×c

Putting the values of a, b, and c, we get

D=(-1)²-4×1×(-1)

D=1-4(-1)

D=1+4=5

The value of D>0.

When the value of the discriminant is greater than 0, its roots are real and unequal.

The equation is

${x}^{6} - x - 1 = 0$

${x}^{6} - x = 1$

$x( {x}^{5} - 1) = 1$

We know that 1 can be obtained as a product of 1 and 1 only.

So,

$x = 1 \: and \: ( {x}^{5} - 1) = 1$

$x = 1 \: and \: {x}^{5} = 2$

Thus, there are only 2 values of x.

Therefore, the equation x6-x-1=0 has exactly two positive real roots.