Question

applying sridhar acharya method solve :
1/x - 2 + 1/x - 3 + 1/x - 4 = 0​

Answer 1

Answer:-

Given:-

$\sf \dfrac{1}{x - 2} + \dfrac{1}{x - 3} + \dfrac{1}{x - 4} =\: 0$

To find:-

• What is the value of x.

Solution:-

$\longmapsto$ $\sf \dfrac{1}{x - 2} + \dfrac{1}{x - 3} + \dfrac{1}{x - 4} =\: 0$

$\implies \sf \dfrac{(x - 3)(x - 4) + (x - 2)(x - 4) + (x - 2)(x - 3)}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} - 4x - 3x + 12 + {x}^{2} - 4x - 2x + 8 + {x}^{2} - 3x - 2x + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} - 7x + 12 + {x}^{2} - 6x + 8 + {x}^{2} - 5x + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} + {x}^{2} + {x}^{2} - 7x - 6x - 5x + 12 + 8 + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{3x}^{2} - 18x + 26}{(x - 2)(x - 3)(x - 4)} =\: 0$

By doing cross multiplication we get,

$\implies \sf {3x}^{2} - 18x + 26 =\: 0$

From this quadratic equation we have to find the value of a , b and c then,

a = 3

b = - 18

c = 26

Now, we have to apply Sridhar Acharya Method, we know that,

$\sf\boxed{\bold{\pink{x =\: \dfrac{- b \pm \sqrt{{b}^{2} - 4ac}}{2a}}}}$

Now, by putting the value of a , b and c we get,

$\implies \sf x =\: \dfrac{-(- 18) \pm \sqrt{{(- 18)}^{2} - 4(3)(26)}}{2(3)}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{324 - 4 \times 78}}{(2 \times 3}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{324 - 312}}{6}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{12}}{6}$

$\implies \sf x =\: 3 \pm \dfrac{1}{\sqrt{3}}$

Then,

$\implies \sf\bold{x =\: 3 + \dfrac{1}{\sqrt{3}}}$

Either,

$\implies \sf\bold{x =\: 3 - \dfrac{1}{\sqrt{3}}}$

$\therefore$ $\sf\boxed{\bold{\green{The\: value\: of\: x\: is\: 3 + \dfrac{1}{\sqrt{3}} ,\: 3 - \dfrac{1}{\sqrt{3}}.}}}$