Question

Solve the following quadratic equations by factorization: $\frac{4}{x}-3=\frac{5}{2x+3}$, x≠0, $-\frac{3}{2}$

Answer 1

SOLUTION :  

Given : (4/x) - 3 = 5/(2x + 3)  

[4 - 3x] /x  = 5/(2x + 3)  

[ By taking LCM]

(4 - 3x) (2x + 3) = 5x

8x +12 - 6x² - 9x = 5x  

-6x² + 8x - 9x - 5x +12 = 0

-6x² - 6x + 12 = 0

-6(x² + x - 2) = 0

x² + x - 2 = 0

x² + 2x  - x - 2 = 0

x(x + 2) - 1(x + 2) = 0

(x -1) (x + 2) = 0

x - 1 = 0   or  x + 2 = 0  

x = 1  or  x = -2

Hence, the roots of the quadratic equation  (4/x) - 3 = 5/(2x + 3)  are 1 & - 2 .

★★ 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

Answer :

The roots of the quadratic equation  (4/x) - 3 = 5/(2x + 3)  are 1 and - 2 .

Step-by-step explanation :

Given - (4/x) - 3 = 5/(2x + 3)  

⇒ [4 - 3x] /x  = 5/(2x + 3)  

By taking the LCM -

⇒ (4 - 3x) (2x + 3) = 5x

⇒ 8x +12 - 6x² - 9x = 5x  

⇒ -6x² + 8x - 9x - 5x +12 = 0

⇒ -6x² - 6x + 12 = 0

⇒ -6(x² + x - 2) = 0

⇒ x² + x - 2 = 0

⇒ x² + 2x  - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x -1) (x + 2) = 0

⇒ x - 1 = 0 OR x + 2 = 0  

⇒ x = 1 OR x = -2

How to solve quadratic equations by factorization :

At first write the given quadratic polynomial as product of the two linear factors by splitting it's middle term.

Then equate both factor equivalent to zero to get the roots of given quadratic equation.