Question

solve –: root ( x-1/3x+2)+2root(3x+2/x-1)=3​

Answer 1

$\large\sf{\sqrt{\frac{x-1}{3x+2}}+2\sqrt{\frac{3x+2}{x-1}}}$

$\sf\blue{Let's\:assume\:that\:}$$\bf\dfrac{x-1}{3x+2}=a^2$

$\sf{Now,the\:equation\:is}$$\sf{\sqrt{a^2}+2\sqrt{\frac{1}{a^2}}=3}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:a+\frac{2}{a}=3}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:\frac{a^2+2}{a}=3}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:a^2+2=3a}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:a^2-3a+2=0}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:a^2-2a-a+2}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:a(a-2)-1(a-2)=0}$

$\:\:\:\:\:\:\:\:\sf{\dashrightarrow\:(a-2)(a-1)=0}$

_______________________

$\bf{Either, \:a - 2 = 0}$

$\bf{:\implies\:a=2}$

$\bf{Or,\:a - 1 = 0}$

$\bf{:⇒ a = 1}$

______________________

${\mathcal{\blue{if\:a = 2\:then\:we\:get,}}}$

$\:\:\:\:\sf\large{\frac{x-1}{3x+2}=2^2}$

$\:\:\:\:\sf\large{⇒ x - 1 = 12x + 8}$

$\:\:\:\:\sf\large{⇒ 11 x = -9}$

$\:\:\:\:\sf\large{⇒ x = -\frac{9}{11}}$

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${\mathcal{\blue{Again\:if\:a = 1\:then\:we\:get,}}}$

$\:\:\:\:\sf\large{\frac{x-1}{3x+2}=1^2}$

$\:\:\:\:\sf\large{⇒ 3x +2 = x - 1}$

$\:\:\:\:\sf\large{⇒3x - x = -1-2}$

$\:\:\:\:\sf\large{⇒ x =-\frac{3}{2}}$

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$\large{ \underline{ \overline{ \mid{ \rm{ \blue{The \:value\:of\:x=-\frac{9}{11},\frac{3}{2}}} \mid}}}}$

Answer 2

Answer:

3x

2

+15x+2

x

2

+5x+1

=2

3(x

2

+5x)+2

x

2

+5x+1

=2

Let x

2

+5x=t

then 3t+2

t+1

=2

4(t+1)=4+9t

2

−12t (squaring both sides)

4t+4=4+9t

2

−12t

9t

2

−16t=0

t=0

x

2

+5x=0

x=0

x=−5

t=

9

16

x

2

+5x=

9

16

9x

2

+45x−16=0

9x

2

+48x−3x−16=0

9x

2

+48x−3x−16=0

3x(3x+16)(3x+16)=0

(3x−1)(3x+16)=0

x=1/3

x=−16/3