Question

Solve the expression given below :-

$\sf {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \dfrac{1\ -\ x}{x}\ =\ \dfrac{17}{x}\ (x \neq 0,2)}$

Answer 1

subtract the sum of -7and-17from the sum of-36and 42

Answer 2

Step-by-step explanation:

Topic :-

Quadratic Equations.

Given :-

We are given that,

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \dfrac{1\ -\ x}{x}\ =\ \dfrac{17}{x}}$

We are also given that,

$\tt : \implies {x \neq 0,2}$

To Find :-

Value of 'x' using Quadratic formula which is commonly known as Shreedhara Acharya's formula.

Solution :-

We solve this question by putting the terms in RHS to LHS. Then we simplify the terms with the same denominator. After that, we take the LCM of the denominators of the two fractions and multiply each fraction with corresponding variables in order to get the denominator of both fractions as LCM. And to get an equation, we add the numerators and then simplify them. Then, we find the roots of the equation using the formula for the roots of the quadratic equation.

$\tt {ax^2\ +\ bx\ +\ c\ =\ 0,\ x\ =\ \dfrac{-b \pm \sqrt{b^2\ -\ 4ac}}{2a}}.$

At last, we simplify the obtained values in order to find the value of x.

Calculations :-

So now let us consider the given equation as,

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \dfrac{1\ -\ x}{x}\ =\ \dfrac{17}{x}}$

Now, let us bring the expression 17/x to the LHS. Then, the equation becomes,

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \dfrac{1\ -\ x}{x}\ =\ \dfrac{17}{x}\ =\ 0}$

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \bigg(\dfrac{1\ -\ x}{x}\ +\ \dfrac{17}{x} \bigg)\ =\ 0}$

Now, the denominators of the fractions inside the bracket are same. So now we can add the numerators. Then, we get,

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \bigg(\dfrac{1\ -\ x\ +\ 17}{x} \bigg)\ =\ 0}$

$\tt : \implies {\dfrac{x\ +\ 3}{x\ -\ 2}\ -\ \dfrac{18\ -\ x}{x}\ =\ 0}$

Taking the LCM of denominators we get LCM as x(x - 2). So now, we simply multiply each fraction accordingly to make the denominator as LCM, we get,

$\tt : \implies {\dfrac{x(x\ +\ 3)}{x(x\ -\ 2)}\ -\ \dfrac{(18\ -\ x)\ (x\ -\ 2)}{x(x\ -\ 2)}\ =\ 0}$

By adding them, we get,

$\tt : \implies {\dfrac{x(x\ +\ 3)\ -\ (18\ -\ x)\ (x\ -\ 2)}{x(x\ -\ 2)}\ =\ 0}$

As we are given that x ≠ 0,2, so we can write the above equation as,

$\tt : \implies {x(x\ +\ 3)\ -\ (18\ -\ x)\ (x\ -\ 2)\ 0}$

On simplifying it we get,

$\tt : \implies {x^2\ +\ 3x\ -\ (18x\ -\ 36\ +\ 2x\ -\ x^2)\ =\ 0}$

$\tt : \implies {x^2\ +\ 3x\ -\ (20x\ -\ 36\ -\ x^2)\ =\ 0}$

Now, we can write it as,

$\tt : \implies {x^2\ +\ 3x\ -\ 20x\ +\ 36\ +\ x^2\ =\ 0}$

$\tt : \implies {2x^2\ -\ 17x\ +\ 36\ =\ 0}$

Now, let us consider the formula for the roots of the quadratic equation ax² + bx + c = 0.

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

Comparing the equation,

$\tt : \implies {2x^2\ -\ 17x\ +\ 36\ =\ 0}$

$\tt : \implies {ax^2\ +\ bx\ +\ c\ =\ 0}$

$\tt : \implies {a\ =\ 2,\ b\ =\ -17,\ c\ =\ 36}$

By substituting the values in the formula we get,

$\tt : \implies {x\ =\ \dfrac{-(-17) \pm \sqrt{(-17)^2\ -\ 4(2)\ (36)}}{2(2)}}$

$\tt : \implies {x\ =\ \dfrac{17 \pm \sqrt{289\ -\ 288}}{4}}$

$\tt : \implies {x\ =\ \dfrac{17 \pm \sqrt{1}}{4}}$

$\tt : \implies {x\ =\ \dfrac{17 \pm 1}{4}}$

Now, we get the value of x as,

$\tt : \implies {x\ =\ \dfrac{17\ +\ 1}{4}\ ,x\ =\ \dfrac{17\ -\ 1}{4}}$

$\tt : \implies {x\ =\ \dfrac{18}{4}\ ,x\ =\ \dfrac{16}{4}}$

$\tt : \implies {x\ =\ \dfrac{9}{2}\ ,x\ =\ 4}$

Answer :-

Hence, the value of x is,

$\tt : \implies {x\ =\ \dfrac{9}{2},\ 4.}$