Question

Solve for x:
x-2 /x-3 +x-4/x-5 =10/3
(quadratic equation)

Answer 1

Answer:

$x = 5.309 \: \: or \: \: x = 4.191$

Explanation:

Given:

$\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}$

To Find:

Value of x

Steps:

Firstly, taking the LCM at LHS

LCM is (x-3)(x-5)

$\frac{(x - 2) \times (x - 5) \: + \: (x - 4) \times (x - 3)}{(x - 3)(x - 5)} = \frac{10}{3}$

Multiplying and performing calculation in numerator and denominator, we get:

$\frac{( {x}^{2} - 5x - 2x + 10) + ( {x}^{2} - 3x - 4x + 12) }{ ({x}^{2} - 3x - 5x + 15) } = \frac{10}{3}$

Opening the bracket and performing calculation, we get:

$\frac{ {x}^{2} + {x}^{2} - 7x - 7x + 10 + 12 }{ {x}^{2} - 8x + 15} = \frac{10}{3}$

Adding and subtracting every like terms , we get:

$\frac{2 {x}^{2} - 14x + 22 }{ {x}^{2} - 8x + 15 } = \frac{10}{3}$

Cross multiplying LHS and RHS, we get:

$6 {x}^{2} - 42x + 66 = 10 {x}^{2} - 80x + 150$

Bringing RHS to LHS and changing signs, we get:

$6 {x}^{2} - 10 {x}^{2} - 42x + 80x + 66 - 150 = 0$

Adding and subtracting every like term, we get:

$- 4 {x}^{2} + 38x - 89 = 0$

Changing the signs so that the square term becomes positive, we get:

$- (4 {x}^{2} - 38x + 89) = 0$

$4 {x}^{2} - 38x + 89 = 0$

Using Sridharacharya Method or Quadratic Formula

Sridharacharya Method (Quadratic Formula)

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

where,

x= roots of equation

a= constant in ax²+bx+c(can be negative or positive)

b= constant in ax²+bx+c(can be negative or positive)

c= constant in ax²+bx+c(can be negative or positive)

Here in this question,

a=4

b=-38

c=89

Substituting the value of a, b and c in the above formula, we get:

$x = \frac{ - ( - 38) \: \pm \: \sqrt{ { - 38}^{2} - 4 \times 4 \times 89 } }{2 \times 4}$

$x = \frac{38 \:\pm \: \sqrt{1444 - 1424} }{8}$

$x = \frac{38 \: \pm \: \sqrt{20} }{8}$

$Value \: of \: \sqrt{20} = 4.472$

$x = \frac{38 \:\pm \: 4.472 }{8}$

$x = \frac{38 + 4.472}{8} \: \: or \: \: x = \frac{ 38 - 4.472}{8}$

$x = 5.309 \: \: or \: \: x = 4.191$

Additional Information:

1) We can solve the question by other method also like Split Method.

2) In Quadratic Formula, the value of a, b and c can be negative if the equation contains negative term or terms.

3) In some questions, if the value of root cannot be negative , then we neglect the negative value of root.

4) We take out the square root of discriminant in quadractic formula by Division Method. b²-4ac is called the discriminant.