Question

The number of distinct real solutions of the equation x2−cosx−1=0 is _____.

Answer 1

Answer:

1 is the answer.......

Answer 2

Answer:

The number of distinct real solutions of the equation x2−cosx−1=0 is 2.

Step-by-step explanation:

Here given equation is

${x}^{2} - cosx - 1 = 0.....(1)$

We know,

For the quadratic equation

$a {x}^{2} + bx + c = 0....(2)$

, the expression

${b}^{2} - 4ac$

is called the discriminant. The value of the discriminant shows how many roots f(x) has: - If ${b}^{2} - 4ac > 0$

then the quadratic function has two distinct real roots. If ${b}^{2} - 4ac = 0$

then the quadratic function has one repeated real root.

We are comparing equation (1) and (2) and get,

$a = 1 \\ b = 0 \\ c = ( - cosx - 1)$

Now,

${b}^{2} - 4ac \\ = {0}^{2} - 4.1.( - cosx - 1) \\ = 4 \cos(x) + 4 \\ = 4(cosx + 1)$

It is clear that

$4(cosx + 1) > 0$

So, we can say

${b}^{2} - 4ac > 0$

So,by rule of quadratic equation,we can say that given equation has two distinct real roots.

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549