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

Given:

Given quadratic equation is $\frac{x\ +\ 3}{x\ +\ 2}\ =\ \frac{3x\ -\ 7}{2x\ -\ 3},\ x\ ≠\ -2,\ \frac{3}{2}$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{x+3}{x+2}=\frac{3x-7}{2x-3}$

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

$2x^2-3x+6x-9=3x^2+6x-7x-14$    (Cross multiplying)

$3x^2-2x^2-x-3x-14+9=0$

$x^2-4x-5=0$

$x^2-5x+x-5=0$

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

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

$x+1=0$ or $x-5=0$

$x=-1$ or $x=5$

The roots of the given quadratic equation are $-1$ and $5$. 

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Updated on: 10-Oct-2022

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