Question

Solve the following quadratic equations by factorization: $\frac{x+3}{x+2} =\frac{3x-7}{2x-3}$

Answer 1

SOLUTION :  

Given :  (x + 3) / (x + 2) = (3x − 7) /(2x − 3)

(x + 3)(2x - 3) = (x + 2)(3x - 7)

2x² -3x + 6x - 9 = 3x² + 6x - 7x  - 14

2x² + 3x - 9 = 3x² - x -14

2x² - 3x² + 3x + x - 9 +14 = 0

-x² + 4x + 5 =0

x² - 4x - 5 = 0  

x² - 5x + x - 5 = 0

x(x - 5) +1(x - 5)= 0

(x - 5) (x + 1 ) = 0

x - 5 = 0 or x + 1= 0

x = 5  or  x= -1

The roots of the above mentioned quadratic equation (x + 3) / (x + 2) = (3x − 7) /(2x − 3) are 5 and -1

★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION METHOD :  

We first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution:-

$= > \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3} \\ \\ = >( 2x - 3)(x + 3) = (x + 2)(3x - 7) \\ \\ = > 2 {x}^{2} + 6x - 3x - 9 = 3 {x}^{2} - 7x + 6x - 14 \\ \\ = > 3 {x}^{2} - 2 {x}^{2} - 7x + 3x - 14 + 9 = 0 \\ \\ = > {x}^{2} - 4x - 5 = 0 \\ \\ = > {x}^{2} - (5 - 1)x - 5 = 0 \\ \\ = > {x}^{2} - 5x + x - 5 = 0 \\ \\ = > x(x - 5) + 1(x - 5) = 0 \\ \\ = > (x - 5)(x + 1) = 0 \\ \\ = > x = 5 \: and \: x = - 1$