Question

Here the some interesting questions Pls solve all !

1 )The number of real roots of the equation x² -3|x| + 2 = 0
Hint :- Answer is 4

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2) If
$x = \sqrt{6 + \sqrt{6 + \sqrt{6} + ... \infty } }$
then the value of x is

Hint :- Answer is 3

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3) The two consecutive positive odd integers such that sum of their squares is 290 are

Hint :- Answer is 11, 13 ​

Answer 1

Answer:

Step-by-step explanation:

Solution :-

1) x² - 3|x| + 2 = 0

We know that |x| = ± x

So we will get two types of quadratic equations :-

x² - 3x + 2 = 0    , x² - 3(-x) + 2 = 0

→ x² - 3x + 2 = 0 , x² + 3x + 2 = 0

So we need know that every quadratic equation has '2' roots

So as we got 2 different quadratic equations , so there will be '4' roots.

∴'x² - 3|x| + 2 = 0' has '4' roots.

2)

$x=\sqrt{6+\sqrt{6+\sqrt{6+..\infty}}}$

Squaring on both sides to the above equation :-

$(x)^2=\left(\sqrt{6+\sqrt{6+\sqrt{6+..\infty}}}\right)^2$

$x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+..\infty}}}$

→ x² = 6 + x

x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

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

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

→ x = -2 , 3

∴ As 'x' value can'nt be negative.

Value of 'x' = 3.

3)

Let , the two consecutive positive odd numbers be :-

x , (x +  2)

According to Question :-

(x²) + (x + 2)² = 290

x² + x² + 4x + 4 = 290

2x² + 4x + 4 - 290 = 0

2x² + 4x - 286 = 0

2(x² + 2x - 143) = 0

x² + 2x - 143 = 0

x² + 13x - 11x - 143 = 0

x(x + 13) - 11(x + 13) = 0

(x + 13) (x - 11) = 0

→ x = 11 , -13

As they given in the question that positive odd numbers , we shouldn't consider that 'x = -13'.

→ x = 11

Another odd number = x + 2 = 11 + 2 = 13

11 , 13.

Answer 2

Given Equation,

$\sf \: f(x) = {x}^{2} - 3 |x| + 2$

Here, |x| = x if x > 0. Thus,

$\sf \: {x}^{2} - 3 x + 2 = 0 \\ \\ \longrightarrow \sf \: x {}^{2} - 2x - x + 2 = 0 \\ \\ \longrightarrow \sf (x - 1)(x - 2) = 0 \\ \\ \longrightarrow \boxed{ \boxed{ \sf x = 1 \: or \: 2}}$

Also, |x| = - x for x < 0. Thus,

$\sf \: {x}^{2} + 3 x + 2 = 0 \\ \\ \longrightarrow \sf \: x {}^{2} + 2x + x + 2 = 0 \\ \\ \longrightarrow \sf (x + 1)(x + 2) = 0 \\ \\ \longrightarrow \boxed{ \boxed{ \sf x = - 1 \: or \: - 2}}$

Roots of the equation are ±1 and ±2. Therefore, total number of real roots is 4.

$\rule{300}{2}$

Given Equation,

$\sf x = \sqrt{6 + \sqrt{6 + \sqrt{6} + ... \infty } }$

The above expression is an infinite nested radical, since the equation is continuous. Substitute x in the radical as -

$\longrightarrow \sf x = \sqrt{6 + x}$

Squaring over both sides,

$\longrightarrow \sf x^2 = 6 + x \\ \\ \longrightarrow \sf x^2 -x - 6 = 0 \\ \\ \longrightarrow \sf x^2 - 3x + 2x -6 = 0 \\ \\ \longrightarrow \sf (x - 3)(x +2) = 0 \\ \\ \longrightarrow \sf x = 3 or -2$

Since, x = -2 is beyond the scope of natural numbers, we shall ignore it.

$\longrightarrow \boxed{\boxed{\sf x = 3}}$

$\rule{300}{2}$

Positive Odd Integers are of the form - 2n - 1, 2n + 1, 2n + 3 and so on

ATQ,

$\sf (2n + 1)^2 + (2n - 1)^2 = 290 \\ \\ \longrightarrow \sf 4n^2 + 4n + 1 + 4n^2 - 4n + 1 = 290 \\ \\ \longrightarrow \sf 8n^2 = 288 \\ \\ \longrightarrow \sf n^2 = 36 \\ \\ \longrightarrow \sf n = 6 \ \ \ \{\because n \in Z^+\}$

Now,

• 2n + 1 = 2(6)+1 = 13
• 2n - 1 = 2(6)-1 = 11

Two consecutive positive odd integers are 11 and 13.

$\rule{300}{2}$

$\rule{300}{2}$