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If the product of two surds is rational, then each of them is called the rationalizing factor of the other. The product of √3+√2 and √3-√2 is 3 which is a rationalizing factor of √3-√2 and vice-versa. √3+√2 and√3-√2 are conjugate to each other. So, for a binomial quadratic surd, its conjugate is the rationalizing factor. Note that √5 is rationalized when it is multiplied by √5 so √5 is rationalizing factor of √5.
Since 2√5 x√5 = 10, then
2√5 and √5 are rationalizing factor of each other.
Let us took at another example.
2√5 x 3√5 =30
∴ 2√5 and 3√5 are rational factors of each other. Thus, surd may have more than one rational factor. But it is better to choose a rational factor that is easy to evaluate.
The equations are which the variables are expressed in terms of surds are called the equations involving surds.\(\sqrt{3x+2}\) +\(\sqrt{3x-11}\) =9 is an example of equation involving surds.
The steps of solving surd equations are as follows:
In the solution of equation involving radical sign, verification must be shown. If any value of the variable doesn't satisfy the given equation, it should be discarded.
Solution:
\(\sqrt{ x}\)-2 = 5
or, \(\sqrt{x}\) = 3+2
or, \(\sqrt{x}\) = 5
Squaring on both sides,
or, \(\sqrt{x}\)^{2} =(5)^{2}
or, x = 25
Checking for x = 25,
\(\sqrt{25}\) - 2 = 3
or, 5 - 2 = 3
or, 3 = 3 (true)
Hence, x = 25.
Solution:
Here, \(\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}\) =a + b \(\sqrt{3}\)
or, \(\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}\) \(\times\) \(\frac{\sqrt3- \sqrt{2}}{\sqrt{3 - \sqrt{2}}}\)= a+b\(\sqrt{3}\)
or, \(\frac{3 \sqrt{3} - 3\sqrt{2} + \sqrt{18} - \sqrt{12}} {3-2}\) = a + b \(\sqrt{3}\)
or, 3\(\sqrt{3} - 3 \sqrt{2} + 3\sqrt{2} - 2 \sqrt{3}\) = a + b \(\sqrt{3}\)
or, \(\sqrt{3}\) = a + b \(\sqrt{3}\)
or, 0 + 1. \(\sqrt{3}\) = a + b \(\sqrt{3}\)
Comparing the coefficients, we get a =0 and b=1
∴ a = 0 and b= 1
Solution:
\(\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} + \sqrt{a^2 -1}}\) \(\times\) \(\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} - \sqrt{a^2 -1}}\)
= \(\frac{(\sqrt{a^2 + 1} - \sqrt{a^2 -1)^2}} {(\sqrt{a^2 + 1})^2 - (\sqrt{a^2 -1})^2}\)
=\(\frac{\sqrt{(a^2 + 1)^2} - 2. \sqrt{a^2 + 1}. \sqrt{a^2 -1} + \sqrt{(a^2 -1)^2}}{(a^2 + 1) - ( a^2 - 1)}\)
=\(\frac{a^2+1-2\sqrt{{(a^2)}^2-{(1)^2}}+a^2 -1}{2}\)
=\(\frac{2a^2-2\sqrt{{a^4}-{1}}}{2}\)
= \(\frac{2( a^2 - \sqrt{a^4 -1)}}{2}\)
= \(a^2 - \sqrt{a^4 - 1}\)
Solution:
\(\frac{1}{\sqrt{7} -1}\)
= \(\frac{1}{\sqrt{7} -1}\) \(\times\) \(\frac{\sqrt{7} +1}{\sqrt{7}+1}\)
= \(\frac{\sqrt{7} +1}{(\sqrt{7})-(1)^2}\)
= \(\frac{\sqrt{7} +1}{7-1}\)
= \(\frac{\sqrt{7} +1}{6}\)
Solve (sqrt{2x+3}) = 5
Find the square root of 13 + 2 (sqrt{30})
Solve: (frac{sqrt{x + 4} + sqrt{x -4}} {sqrt{ x +4 } - sqrt{x -4}})=2
Rationalize the denominator
(frac{1}{sqrt2 -1})
Solve:
(sqrt{3x +1}) = x -3
Solve:
x + (sqrt{x}) - 6 = 0
Solve:
(sqrt{x}) - 2 = 3
What is the rationalizing factor of
(sqrt{x + a})
ASK ANY QUESTION ON Rationalization of Surds
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