Question

Solve Que no.24 ASAP!! :)​

Answer 1

$\frac{m}{n} x {}^{2} + \frac{n}{m} = 1 - 2x \\ \\ \implies \: m {}^{2}x² +n² = mn - 2mnx \\ \\ \implies m ²x²+2mnx+n² −mn=0$

Here, a = m², b = 2mn, c = n²- mn

➯ By using the Quadratic Formula Or Shri Dharacharya Formula we have

$</p><p> \Large x = \frac{ - b \: \pm \: \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ \implies x= \frac{ -2mn \: \pm \: \sqrt{(2mn){}^{2} - 4(m²)(n²-mn)} }{2m²} \\ \\ \implies</p><p>x= \frac{ -2mn \: \pm \: \sqrt{4m²n² - 4m²n²+4m {}^{3} n} }{2a} \\ \\ \implies x= \frac{ -2mn \: \pm \: 2m\sqrt{mn} }{2m²} \\ \\ \implies x= \frac{ - n \: \pm \sqrt{mn} }{m} </p><p></p><p></p><p>$

Answer 2

Step-by-step explanation:

The problem is about finding the values of x (that is factorisation). In order to solve this problem, firstly we have to change the given equation in the form of quadratic equation. General form of quadratic equation is $\tt a{x}^{2} + bx + c = 0$. We will solve this question by applying quadratic formula.

So let's proceed by simplifying the equation!

$\tt{: \implies \dfrac{m}{n} x^{2} + \dfrac{ n}{m} = 1 - 2x}$

Taking LCM

$\tt{: \implies \dfrac{m ^{2} {x}^{2} + {n}^{2} }{mn} = 1 - 2x}$

$\tt{: \implies m ^{2} {x}^{2} + {n}^{2} =mn( 1 - 2x)}$

$\tt{: \implies m ^{2} {x}^{2} + {n}^{2} =mn- 2mnx}$

$\tt{: \implies m ^{2} {x}^{2} + 2mnx + ({n}^{2} - mn) = 0 }$

Now we can see that this equation is in the general form of a quadratic equation where $\tt a = {m}^{2}$ , $\tt b = 2mn$ and $\tt c = (n^{2}-mn)$.

Now we have quadratic formula :-

$\boxed{ \tt \purple{ : \implies x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

Substitute the values of a, b and c

$\tt{: \implies x = \dfrac{ - (2mn) \pm \sqrt{ {(2mn)}^{2} - 4( {m}^{2} )( {n}^{2} - mn ) } }{2a}}$

$\tt{: \implies x = \dfrac{ - 2mn \pm \sqrt{ {4m^{2} n^{2} } - 4( {m}^{2} )( {n}^{2} - mn ) } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2mn \pm \sqrt{ {4m^{2} n^{2} } - 4( {m}^{2} )( {n}^{2} - mn ) } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2mn \pm \sqrt{ {4m^{2} n^{2} } -4 {m}^{2} {n}^{2} + 4 {m}^{3}n } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2mn \pm \sqrt{ 4 {m}^{3}n } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2mn \pm 2m\sqrt{ mn } }{2 {m}^{2} }}$

‎ ‎ ‎

For negative :-

$\tt{: \implies x = \dfrac{ - 2mn - 2m\sqrt{ mn } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2m(n + \sqrt{ mn }) }{2 {m}^{2} }}$

$\pink{\boxed{ \tt{: \implies x = \dfrac{ - \sqrt{ mn } - n}{m }}}}$

‎ ‎ ‎

For positive :-

$\tt{: \implies x = \dfrac{ - 2mn + 2m\sqrt{ mn } }{2 {m}^{2} }}$

$\tt{: \implies x = \dfrac{ - 2m(n - \sqrt{ mn } )}{2 {m}^{2} }}$

$\green{ \boxed{ \tt{: \implies x = \dfrac{ \sqrt{ mn } - n }{m }} }}$

These are the required answers.