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The sum and difference of two simple quadratic surds are said to be conjugate surds to each other.
Conjugate surds are also known as complementary surds.
Thus, the sum and the difference of two simple quadratic surds 4β7and β2 are 4β7 + β2 and 4β7 - β2 respectively. Therefore, two surds (4β7 + β2) and (4β7 - β2) are conjugate to each other.
Similarly, two surds (-2β5 + β3) and (-2β5 - β3) are conjugate to each other.
In general, two binomial quadratic surds (xβa + yβb) and (xβa - yβb) are conjugate to each other.
In complex or binominal surds, if sum of two quadratic surds or a quadratic surd and a rational number is multiplied with difference of those two quadratic surds or quadratic surd and rational number, then rational number under root of surd is get squared off and it becomes a rational number as product of sum and difference of two numbers is difference of the square of the two numbers.
a2βb2=(a+b)(aβb).
The sum and difference of two quadratic surds is called as conjugate to each other. For example βx = a and βy = b, a and b are two quadratic surds, if (a + b) or (βx+βy) is multiplied with (a - b) or (βxββy), the result will (βx)2 - (βy)2 or (x - y) which is rational number. Here (βx+βy) and (βxββy) are conjugate surds to each other and the process is called as rationalization of surds as the result becomes a rational number. This process is used for fraction expression of complex surds, where the denominator needs to converted to a rational number eliminating the roots of surds, conjugate surds multiplied to both numerator and denominator and denominator becomes rational.
Like for example, if simplification of the complex surd 6β3β1 is to be done, denominator β3β1 is to be converted to a rational number. If a = β3 and b = 1, then denominator is (a-b), if we multiply (a + b) or β3+1, it will a2βb2 and β3 will be squared off.
6β3β1
= 6(β3+1)(β3β1)(β3+1)
= 6(β3+1)3β1
= 6(β3+1)2
= 2(\sqrt{3} + 1).
In the above example β3+1 is used as rationalizing factor which is a conjugate to β3β1.
Note:
1. Since 3 + β5 = β9 + β5 and surd conjugate to β9 + β5 is β9 - β5, hence it is evident that surds 3 + β5 and 3 - β5 are conjugate to each other.
In general, surds (a + xβb) and (a - xβb) are complementary to each other.
2. The product of two binomial quadratic surds is always rational.
For example,
(βm + βn)(βm - βn) = (βm)^2 - (βn)^2 = m - n, which is rational.
Here are some examples of conjugates in the following table.
(β2+β3) (β5+β3) β2+1 (4β2+2β3) (βx+y) (aβx+bβy) |
(β2ββ3) (β5ββ3) β2β1 (4β2β2β3) (βxβy) (aβxβbβy) |
Problems on conjugate surds:
1. Find the conjugates of the following surds.
(β5+β7), (4β11β3β7), 3β17+19, (aβbβbβa).
Solution:
Given Surds (β5+β7) (4β11β3β7) 3β17+19 (aβbβbβa) |
Conjugate (β5ββ7) (4β11+3β7) 3β17β19 (aβb+bβa) |
2. Simplify the surd 2β5β12β5+1 by using conjugate surd.
Solution:
= 2β5β12β5+1
As the denominator is 2β5+1, for rationalization of the surd, we need to multiply both numerator and denominator by the conjugate surd 2β5β1.
= (2β5β1)(2β5β1)(2β5+1)(2β5β1)
= (2β5β1)25β1β¦.. as we know (a+b)(aβb)=a2βb2
= ((2β5)2β2Γβ5+12)4
= 5β2β5+14
= 6β2β54
= 2(3ββ5)4
= 3ββ52
3. Rationalize the surd β2βxββ2.
Solution:
β2βxββ2
As the denominator is (βxββ2), the conjugate surd is (βx+β2), we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= (β2)(βx+β2)(βxββ2)(βx+β2)
= β2x+2xβ2.
4. Rationalize the surd β52β7β3β5.
Solution:
β52β7β3β5
As the denominator is (2β7β3β5), the conjugate surd is (2β7+3β5), we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= β5Γ(2β7+3β5)(2β7β3β5)(2β7+3β5)
= 2β35+3Γ5(2β7)2β(3β5)2
= 2β35+154Γ7β9Γ5
= 2β35+1528β45
= β(2β35+15)17
β Surds
11 and 12 Grade Math
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