Math, asked by Anonymous, 1 month ago

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 ​

Answers

Answered by sharanyalanka7
102

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.


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Answered by Anonymous
108

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}


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