Question

Solve the following inequality with the help of wavy curve method:-

$\bigstar \boxed{ \sf \frac{ \sqrt{2 {x}^{2} + 15x - 17 } }{10 - x} \geqslant 0 } \bigstar$

Have a great answering time :D​

Answer 1

Given :-

√2x² + 15x - 17/10 - x ≥ 0

To Find :-

The solution of the given inequality with the help of wavy curve method .

Solution :-

At first , you should note that x can't be equal to 10 . Because if x becomes 10 then. , the denominator becomes 0 . So to avoid this we will not take x as 10 .

=> √2x² + 15x - 17/10 - x ≥ 0

=> Multiplying both sides by √2x² + 15x - 17 we get ,

=> 2x² + 15x - 17/10 - x ≥ 0

=> 2x² - 2x + 17x - 17 / 10 - x ≥ 0

=> 2x ( x - 1 ) + 17 ( x - 1 ) / 10 - x ≥ 0

=> ( 2x + 17 ) ( x - 1 ) / 10 - x ≥ 0

=> Multiplying both sides by ( 10 - x )² we get ,

=> ( 10 - x ) ( 2x + 17 ) ( x - 1 ) ≥ 0

Now , By equating all terms to 0 we get , -17/2. ,10 and 1 .

=> Now Mark 10 , 1 and -17/2 On the number line

=> By wavy curve method , the possible values of "x" are such that :-

• -17/2 ≥ x
• x ≥ 1
• x > 10
• x ≠ 10

Since. ,by the above conditions for value of "x" The solution set is given by :-

=> ( - ∞ , -17/2 ] U [ 1 , ∞ )

=> But as here 10 lies in the given solution set so we use difference of set to avoid "x" as 10 . Henceforth the required solution set is :-

=> ( - ∞ , -17/2 ] U [ 1 , ∞ ) - { 10 }

You should note that wavy curve method is also known as Method of intervals .

Answer 2

Given Expression :-

$\\ \quad \bullet \quad \sf \bigg(x - \dfrac{1}{x} \bigg) ^{ \dfrac{1}{2} } + \bigg(1 - \dfrac{1}{x} \bigg)^{ \dfrac{1}{2} }=x$

This expression can be written as follows :-

$\\ \quad \bullet \quad \underline{ \boxed{\sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} + \sqrt{\bigg(1 - \dfrac{1}{x} \bigg)}=x} } \frak{ \: - - - (i)}$

Solution :-

Multiply the Above eq (i) with the conjugates of LHS

$\\ \quad \longrightarrow \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} + \sqrt{\bigg(1 - \dfrac{1}{x} \bigg)} \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)} =x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} ^{2} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}^{2} = x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \bigg(x - \dfrac{1}{x} \bigg)- \bigg( 1 - \dfrac{1}{x} \bigg) = x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \frac{x - 1}{x} = \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf 1 - \frac{1}{x} = \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \underline{\boxed{ \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)} = 1 - \frac{1}{x} }} \frak{ \: - - - (ii)}$

Now, by adding equations (i) and (ii) we get :-

$\\ \quad \longrightarrow \quad \underline{\boxed{ \sf 2\sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = x - \frac{1}{x} + 1}} \frak{ \: - - - (iii)}$

Assume the following :-

$\quad \mapsto \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = a$

$\quad \therefore \quad \sf \bigg(x - \dfrac{1}{x} \bigg) = {a}^{2}$

Now, substituting the above values in equation (iii) , we get

$\\ \quad \longrightarrow \quad \sf 2a = {a}^{2} + 1$

$\\ \quad \longrightarrow \quad \sf {a}^{2} - 2a + 1 = 0$

$\quad \longrightarrow \quad \sf {(a - 1)}^{2} = 0$

$\\ \quad \longrightarrow \quad \boxed{\sf a = \pm 1}$

Now,

$\\ \quad : \implies \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = a$

$\\ \quad : \implies \quad \sf \bigg(x - \dfrac{1}{x} \bigg) = 1$

$\\ \quad : \implies \quad \sf {x}^{2} - 1 = x$

$\\ \quad : \implies \quad \sf {x}^{2} - x - 1 = 0$

$\\ \quad : \implies \quad \sf x = \frac{ - ( - 1) \pm \sqrt{ { ( - 1)}^{2} - 4( - 1)(1)} }{2}$

$\\ \quad \therefore \quad \boxed{\frak{ x = \frac{ 1 \pm \sqrt{5} }{2}}} \bigstar$