The equation is .
Solve the equation.
\ (Original equation)
(Square on each side)
(Cancel square and root terms)
(Apply differences of squares:
)
(Evaluate power
)
(Subtract
from each side)
(Apply additive inverse property:
)
(Subtraction:
)
(Add
to each side)
(Apply additive inverse property:
)
(Apply additive inverse property:
)
(Factors of equation)
(Factors of group terms)
(Take out common terms)
and
(Separate solutions)
and
(Simplify)
Therefore, the value of are
and
.
Check the solutions for -values:
Case(i):
(Original equation)
(Substitute
in the equation)
(Multiply)
(Subtract:
and
)
(Substitute:
)
Therefore the value of is
does not satisified the equation.
Case(ii):
(Original equation)
(Substitute
in the equation)
(Multiply)
(Subtract:
and
)
(Substitute:
)
Therefore, the value of is
satisified the equation.
The value of is
.