Question

find the roots 1/x+1+1/x+2=4/x+4
x≠1,-2&-4​

Answer 1

Step-by-step explanation:

$\frac{1}{x + 1} + \frac{1}{x + 2} = \frac{4}{x + 4}$

$\frac{(x + 2) + (x + 1)}{(x + 1)(x + 2)} = \frac{4}{x + 4}$

$\frac{2x + 3}{ {x}^{2} + 3x + 2} = \frac{4}{x + 4}$

(x+4)(2x+3) = 4(x²+3x+2)

2x²+3x+8x+12 = 4x²+12x+8

2x²+11x+12 = 4x²+12x+8

2x²+x-4 =0

by quadratic formula

Answer 2

Question :-

Find the roots of :

$\to \tt \frac{1}{x + 1} + \frac{1}{x + 2} = \frac{4}{x + 4} \\ \\ \tt (x ≠ - 1,-2 \: and \: -4 )$

Answer :-

$\tt \to \boxed{ \tt x = \frac{ - 1 \pm \sqrt{33} }{4} } \:$

Used formula :-

If any Quadratic polynomial is given as

→ f(x) = ax² + bx + c = 0

So ,shri dharacharya principle is -

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

Solution :-

We have ,

$\to \tt \frac{1}{x + 1} + \frac{1}{x + 2} = \frac{4}{x + 4} \: \\ \\ \sf taking \: lcm \\ \\ \to \tt \frac{(x + 2) + (x + 1)}{(x + 1) (x + 2)} = \frac{4}{x + 4} \\ \\ \to \tt \: \frac{2x + 3}{(x + 1)(x + 2)} = \frac{4}{x + 4} \\ \\ \sf cross \: multiply \: \\ \\ \to \tt (2x + 3)(x + 4) = 4(x + 1)(x + 2) \\ \\ \to \tt \: 2 {x}^{2} + 8x + 3x + 12 \\ \: \: \: \: \: \: \tt = 4( {x}^{2} + 2x + x + 2) \\ \\ \to \tt \: 2 {x}^{2} + 11x + 12 = 4 {x}^{2} + 12x + 8 \\ \\ \sf \: seprate \: all \: of \: these \: \\ \\ \to \tt 2 {x}^{2} + x - 4 = 0 \\ \\ \sf apply \: shri \: dharacharya \: principle$

$\to \tt x = \frac{ - 1 \pm \sqrt{ {1}^{2} - 4 \times 2 \times ( - 4)} }{2 \times 2} \\ \\ \tt \to \: x = \frac{ - 1 \pm \sqrt{1 + 32} }{4} \\ \\ \tt \to \boxed{ \tt x = \frac{ - 1 \pm \sqrt{33} }{4} }$