Question

if square root of x-1, square root of x+1, +1 =0,then 4x=___​

Answer 1

Answer:

Correct option is A)

x−1

−

x+1

+1=0

x−1

−

x+1

=−1

take square both side

[

(x−1)

−

(x+1)

]

2

=(−1)

2

apply (a−b)

2

=a

2

+b

2

−2ab

(x−1)+(x+1)−2

(x−1)(x+1)

=1

2x−1=2

x

2

−1

again take square both side

(2x−1)

2

=(2

x

2

−1

)

2

4x

2

+1−4x=4(x

2

−1)

4x

2

+1−4x=4x

2

−4

1+4=4x

4x=5

Answer 2

The Answer is: 4x = 5

Explanation:

$\sqrt{x-1}-\sqrt{x+1}+1=0$

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

Taking square on both sides

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

By using: $(a-b)^{2} =a^{2}+b^{2} -2ab$

Here, a=$\sqrt{x-1}$ and b=$\sqrt{x+1}$

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

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

Again Squaring on both sides

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

$4x^{2} +1-4x=4(x^{2} -1)$

$4x^{2} +1-4x=4x^{2} -4$

$4x^{2} -4x^{2} + 1-4x=-4$

$-4x=-4-1$

$-4x=-5$

 $4x=5$