A surd is said to be in its simplest form if the number under the root sign has no perfect square as a factor. For eg. √12 = √4 x √3 = 2√3. There are few questions on which questions are asked in the exam. There are direct trick and also very useful for time bound exams.
Note: n√a is a surd of nth term and it is irrational, whereas 'a' is a rational number
(b) n√ab = n√a x n√b
(c) n√(a/b) = n√a/n√b
(d) (n√a)n = a
(e) m√n√a = mn√a
(f) (n√a)m = n√am
How it derived:
- Take Type(I): Here, the equation is √6+√6+√6+√6+....infinity
Hence, put this equation = x
Step-1:
√6+√6+√6+... infinity = x
Squaring both sides
6+√6+√6+√6+....infinity = x2
Step-2:
Now, using step 1,
6+x = x2 => x2-x-6 = 0 {solve this quadratic equation}
you will get..... (x-2)(x-3)
for Answer:
- you get, x= √6+√6+√6+.... infinity = 2,3
to choose the correct answer, see the sign used. "+" sign has been used. Hence, answer will be 3.
If "-" sign would be there, √6-√6-√6-....infinity, answer will be 2 {as in the case of type (II) question}
Note: n√a is a surd of nth term and it is irrational, whereas 'a' is a rational number
Law of Surds:
(a) n√a = a1/n(b) n√ab = n√a x n√b
(c) n√(a/b) = n√a/n√b
(d) (n√a)n = a
(e) m√n√a = mn√a
(f) (n√a)m = n√am
How it derived:
- Take Type(I): Here, the equation is √6+√6+√6+√6+....infinity
Hence, put this equation = x
Step-1:
√6+√6+√6+... infinity = x
Squaring both sides
6+√6+√6+√6+....infinity = x2
Step-2:
Now, using step 1,
6+x = x2 => x2-x-6 = 0 {solve this quadratic equation}
you will get..... (x-2)(x-3)
for Answer:
- you get, x= √6+√6+√6+.... infinity = 2,3
to choose the correct answer, see the sign used. "+" sign has been used. Hence, answer will be 3.
If "-" sign would be there, √6-√6-√6-....infinity, answer will be 2 {as in the case of type (II) question}
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