Question

Solve the following equations :
(i) $\sqrt{x+1}$ + $\sqrt{x-1}$ = 1
(ii) $\sqrt{x+1}$-$\sqrt{x-1}$=1.

Answer 1

(i) Squaring both members of the given equation, we get
x + 1 + x -1 + 2$\sqrt{x^2-1}$ = 1
or 2$\sqrt{x^2-1}$=1-2x
or 4(x${}^2$-1) = 1- 4x + 4x${}^2$ $\therefore$ x = 5/4.
Since x=5 / 4 does not satisfy the given equation, the equation has no roots.
Important Note : Squaring an equation generally leads to an equation not equivalent to. the given equation and this new equation in addition to the roots of the given equation may have other roots different from them (the so-called extraneous' roots). Hence it is necessary to check, by substitution, whether 5 / 4 is really the root of the original equation. The check shows that 5/4 does not satisfy the original equation. Another important point to be noted is that only principal values of the roots are to be considered. Thus if n is odd, the symbol $\sqrt[n]{na}$ is understood as the only real number whose nth power is equal to a. In this case a may be positive or negative. For example, principal value of $\sqrt[3]{38}$ is 2 and the principal value of $\sqrt[3]{3-27}$ is - 3. If n is even, the symbol $\sqrt[n]{na}$ is understood as the only positive number nth power of which is equal to a Here, necessarily a $\ge$ 0. Under these conditions, for example, we have
$\sqrt{a^2}$ = a if a>0 and $\sqrt{a^2}$ =-a if a<0.
Thus the principal value of $\sqrt{4}$ is 2, the principal value of $\sqrt[4]{481}$ is 3 and the principal value of $\sqrt{(-3{)}^2}$ is -(-3) = 3 etc.
Alternative.l + m = 1. But l${}^2$ - m${}^2$ = 2
$\therefore$ l-m = 2 $\therefore$ 2l = 3
or l${}^2$= x + l = 9/4 $\therefore$ x = 5/4
But this value does not satisfy as it will make 2 = 1
(ii) Proceeding as in part (i), we find that x = 5 / 4 is the root of the equation.
Here check shows that x = 5 / 4 is really a root of the equation.