Question

Given that x is real than the solution set of the equation $\sqrt {x - 1} + \sqrt {x + 1} = 1$ is,
$\left( a \right)$ No solution
$\left( b \right)$ Unique solution
$\left( c \right)$ 2 solutions
$\left( d \right)$ None of these

Answer 1

In this particular question use the concept that if x belongs to real numbers so the solution of square root is always real corresponding to the values of x, so use this concept to reach the solution of the question.

**Complete step by step answer:
Given equation –
$\sqrt {x - 1} + \sqrt {x + 1} = 1$
Now we have to find the set of solutions for x, where x belongs to the real number.
So take $\sqrt {x + 1}$ on the R.H.S side of the equation we have,
$\Rightarrow \sqrt {x - 1} = 1 - \sqrt {x + 1}$
Now apply square root on both sides we have,
$\Rightarrow {\left( {\sqrt {x - 1} } \right)^2} = {\left( {1 - \sqrt {x + 1} } \right)^2}$
Now expand the square using the property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we have,
$\Rightarrow x - 1 = 1 + x + 1 - 2\sqrt {x + 1}$
Now cancel out the same terms from both sides of the above equation we have,
$\Rightarrow - 1 = 2 - 2\sqrt {x + 1}$
$\Rightarrow 2\sqrt {x + 1} = 3$
Now again apply square root on both sides we have,
$\Rightarrow {\left( {2\sqrt {x + 1} } \right)^2} = {3^2}$
Now simplify this we have,
$\Rightarrow 4\left( {x + 1} \right) = 9$
Now divide by 4 throughout we have,
$\Rightarrow \left( {x + 1} \right) = \dfrac{9}{4}$
Now again simplify this we have,
$\Rightarrow x = \dfrac{9}{4} - 1$
$\Rightarrow x = \dfrac{{9 - 4}}{4} = \dfrac{5}{4}$
Now it is given that x is real
$\Rightarrow x \geqslant 1$, otherwise $\sqrt {x - 1}$ will lead to a complex number. I.e. suppose x = 0.9
Therefore, $\sqrt {x - 1} = \sqrt {0.9 - 1} = \sqrt { - 0.1} = i\sqrt {0.1}$, $\left[ {\because i = \sqrt { - 1} } \right]$.
So x should be greater than or equal to one.
So the solution which we calculated satisfies the above condition.
But when we substitute this value in the L.H.S part we get,
$\Rightarrow \sqrt {x - 1} + \sqrt {x + 1} = \sqrt {\dfrac{5}{4} - 1} + \sqrt {\dfrac{5}{4} + 1} = \sqrt {\dfrac{1}{4}} + \sqrt {\dfrac{9}{4}}$
$\Rightarrow \sqrt {x - 1} + \sqrt {x + 1} = \pm \left( {\dfrac{1}{2} + \dfrac{3}{2}} \right) = \pm \dfrac{4}{2} = \pm 2$
Which is not equal to R.H.S.
$\Rightarrow {\text{ }} \ne R.H.S$
So the solution of the given equation is not possible.
So this is the required answer.
Hence option (A) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that after solving the equation as above solved, substitute the values in the L.H.S part of the equation and simplify as above simplified if the resultant equal to the R.H.S part of the equation then the solution is the required solution of the equation and if not then there is no solution of the equation.