Question

1/x+1+2/x+2=4/x+4ko hal kijiye​

Answer 1

Answer:

Note that the given equation is not in the standard form of a quadratic equation.

Consider

x+1

1

+

x+2

2

=

x+4

4

That is,

x+1

1

=2[

x+4

2

−

x+2

1

]=2[

(x+4)(x+2)

2x+4−x−4

]

x+2

1

=2[

(x+2)(x+4)

x

]

x

2

+6x+8=2x

2

+2x

Thus, we have x

2

−4x−8=0, which is a quadratic equation.

(The above equation can also be obtained by taking LCM )

Using the quadratic formula we get,

x=

2(1)

4±

16−4(1)(−8)

=

2

4±

48

Thus, x=2+2

3

or 2−2

3

Hence, the solution set is {2−2

3

,2+2

3

}

Answer 2

$\bold{x=(2 \pm \sqrt{3})}x=(2±3){\bf{ is\: the \:value\: of\: x \:for}} \bold{\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}}$

Given:

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

To find:

• Value of x =?

Solution:

• To find the value of x, we simplify the following solution:

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

$\frac{1}{x+1}+\frac{2}{x+2}-\frac{4}{x+4}=0$

• Taking the LCM of (x+1)(x+2)(x+4) we find the value of the question in simple form:

$\frac{(x+2)(x+4)+2(x+1)(x+4)-4(x+1)(x+2)}{(x+1)(x+2)(x+4)}$=0

$(x+1)(x+2)(x+4)$

• Transferring (x+1)(x+2)(x+4) over to zero to reduce it to zero.

$(x+2)(x+4)+2(x+1)(x+4)−4(x+1)(x+2)$

${\bf{(x+2)(x+4)+2(x+1)(x+4)-4(x+1)(x+2)=0}}$

• Multiplying and subtracting the extra parts

$\begin{gathered}\begin{array}{l}{x^{2}+6 x+8+2 x^{2}+10 x+8=4 x^{2}+12 x+8} \\ {x^{2}-4 x-8=0}\end{array}\end{gathered}$

• As we can see that the equation cannot be factorize using rational number. Hence the values of the roots are:

$\begin{gathered}\begin{array}{l}{\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {x^{2}-4 x-8=0}\end{array}\end{gathered}$

$\purple{\bf{The \:value\: of \:a= 1, b=-4\:and\:c=-8}}$

$\begin{gathered}\begin{array}{l}{=\frac{-(-4) \pm \sqrt{16-4(1)(-8)}}{2}} \\ {(x-(2+\sqrt{3}))(x+(2-\sqrt{3}))}\end{array}\end{gathered}$

Therefore, the value of x is $\bold{x=(2 \pm \sqrt{3})}$