Question

1/x+5+3/4(3x+1)=1/x+2, x≠-5,x≠-2,x≠-1/3,Solve the given quadratic equation using factorization.

Answer 1

HELLO DEAR,

we know in factorization method we have to splitting middle term in the form of two numbers. in summation of two no. Gives middle no. And in multiplication gives multiple of another two no.

So, given polynomial is .
$\bold{\sf{\frac{1}{x + 5} + \frac{3}{4(3x + 1)} = \frac{1}{x + 2}}}$

$\bold{\sf{\frac{4(3x + 1) + 3(x + 5)}{(x + 5)(12x + 4)} = \frac{1}{x + 2}}}$

$\bold{\sf{\frac{15x + 19}{(12x^2 + 4x + 60x + 20)} = \frac{1}{x + 2}}}$

$\bold{\sf{(15x + 19)(x + 2) = (12x^2 + 64x + 20)}}$

$\bold{\sf{15x^2 + 49x + 38 = 12x^2 + 64x + 20}}$

$\bold{\sf{15x^2 - 12x^2 + 49x - 64x + 38 - 20 = 0}}$

$\bold{\sf{3x^2 - 15x + 18 = 0}}$

$\bold{\sf{x^2 - 5x + 6 = 0}}$

$\bold{\sf{x^2 - 3x - 2x + 6 = 0}}$

$\bold{\sf{x(x - 3) - 2(x - 3) = 0}}$

$\bold{\sf{(x - 2)(x - 3) = 0}}$

$\sf{\large{x = 2 , x = 3}}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi ,

It is given that ,

1/(x + 5 ) + 3/4( 3x + 1 ) = 1/( x + 2 )

=> [4(3x+1)+3(x+5)]/[(x+5)4(3x+1)]=1/(x+2)

=>[12x+4+3x+15]/[12x²+64x+20]=1/(x+2)

=>[(15x+19)(x+2)] = 12x² + 64x + 20

=> 15x² + 30x + 19x + 38 - 12x² - 64x - 20 = 0

=> 3x² - 15x + 18 = 0

=> 3( x² - 5x + 6 ) = 0

x² - 5x + 6 = 0

Splitting the middle term ,

x² -2x - 3x + 6 = 0

x( x -2 ) - 3( x - 2 ) = 0

( x - 2 )( x - 3 ) = 0

x - 2 = 0 or x - 3 = 0

Therefore ,

x = 2 or x = 3

I hope this helps you.

: )