Question

the given equation is given then find the solution of this equation
$\sqrt{5x - 1} + \sqrt{x - 1} = 2$
​

Answer 1

Solution:

$\sqrt{5x - 1} + \sqrt{x - 1} = 2$

Squaring on both sides:

$5x - 1 = 4 + x - 1 - 4 \sqrt{x - 1}$

$x - 1 = - \sqrt{x - 1}$

Squaring on both sides:

${(x - 1)}^{2} = ({ - \sqrt{x - 1} })^{2}$

$= > {x}^{2} - 2x + 1 = x - 1$

$= > {x}^{2} - 3x + 1 + 1 = 0$

$= > {x}^{2} - 3x + 2 = 0$

$= > (x - 2)(x - 1) = 0$

$Therefore, \: x = 2 \: or \: x = 1$

But x = 2 is not satisfying the given equation.

[So, x = 1 is the only solution.]

Answer 2

olution:

5x−1+x−1=2\sqrt{5x - 1} + \sqrt{x - 1} = 2

5x−1

+

x−1

=2

Squaring on both sides:

5x−1=4+x−1−4x−15x - 1 = 4 + x - 1 - 4 \sqrt{x - 1}5x−1=4+x−1−4

x−1

x−1=−x−1x - 1 = - \sqrt{x - 1}x−1=−

x−1

Squaring on both sides:

(x−1)2=(−x−1)2{(x - 1)}^{2} = ({ - \sqrt{x - 1} })^{2}(x−1)

2

=(−

x−1

)

2

=>x2−2x+1=x−1= > {x}^{2} - 2x + 1 = x - 1=>x

2

−2x+1=x−1

=>x2−3x+1+1=0= > {x}^{2} - 3x + 1 + 1 = 0=>x

2

−3x+1+1=0

=>x2−3x+2=0= > {x}^{2} - 3x + 2 = 0=>x

2

−3x+2=0

=>(x−2)(x−1)=0= > (x - 2)(x - 1) = 0=>(x−2)(x−1)=0

Therefore,x=2orx=1Therefore, \: x = 2 \: or \: x = 1Therefore,x=2orx=1

But x = 2 is not satisfying the given equation.

[So, x = 1 is the only solution.]