Question

find the roots of Q.E using quadratic formula 1/ x+ 1 + 2/ x+2 = 4/ x+4​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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

Now,

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

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

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

$= > (3x + 4)(x + 4) = 4( {x}^{2} + 3x + 2)$

$= > 3 {x}^{2} + 12x + 4x + 16 = 4 {x}^{2} + 12x + 8$

$= > 3 {x}^{2} + 16x + 16 = 4 {x}^{2} + 12x + 8$

$= > 0 = 4 {x}^{2} - 3 {x}^{2} + 12x - 16x + 8 - 16$

$= > {x}^{2} - 4x - 8 = 0$

Now, By using quadratic formula,

That is,

$x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a}$

$= > x = \frac{ - ( - 4) \binom{ + }{ - } \sqrt{ {( - 4)}^{2} - 4.1.( - 8) } }{2.1}$

$= > x = \frac{4 \binom{ + }{ - } \sqrt{16 + 32} }{2}$

$= >x = \frac{4 \binom{ + }{ - } \sqrt{48} }{2}$

$= > x = \frac{4 \binom{ + }{ - }4 \sqrt{3} }{2}$

$= >x = \frac{2(2 \binom{ + }{ - }2 \sqrt{3}) }{2}$

$= > x = 2 \binom{ + }{ - } 2 \sqrt{3}$

Therefore, Roots of this Q.E. using quadratic equation are:

$2 + 2 \sqrt{3}$

And

$2 - 2 \sqrt{3}$

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