When a conjugate is multiplied by the original number it removes the surd (or in the case of complex numbers, the complex conjugate removes the imaginary part of the number); this is usually done with a fraction with a surd (or complex number) in a denominator - both numerator and denominator are multiplied by the conjugate. To do this replace a plus by a minus, or a minus by a plus.
Thus (5 - √3) would become either (5 + √3) or (-5 - √3), though it is usual to change the sign of the surd. When multiplied together, they become the difference of two squares; depending upon which sign is changed the result is either positive or negative:
(5 - √3)(5 + √3) = 5² - (√3)² = 25 - 3 = 22
(5 - √3)(-5 - √3) = (5 - √3)((5 + √3) × -1) = ((5 - √3)(5 + √3)) × -1 = (5² - (√3)²) × -1 = 22 × -1 = -22.
The conjugate of a complex number is obtained by changing the sign of the imaginary part. In this case, the complex number is 5 - β3. Therefore, the conjugate of 5 - β3 is 5 + β3. The conjugate pairs have the same real part but opposite imaginary parts, which helps simplify operations involving complex numbers.
a+ square root of b has a conjugate a- square root of b and this is used rationalize the denominator when it contains a square root. If we want to multiply 5 x square root of 10 by something to get rid of the radical you can multiply it by square root of 10. But if we look at 5x( square root of 10 as ) 0+ 5x square root of 10 then the conjugate would be -5x square root of 10
5
Square root of 6
√3 = 1.73205081√5 = 2.23606798
You can get a decimal approximation with a calculator, with Excel, etc. But if you want to keep it as a square root, the "standard form" is considered to be one that has no square roots in the denominator. In this case, to get rid of the square root in the denominator, you multiply both numerator and denominator by the square root of 5, with the following result: 3 / root(5) = 3 root(5) / root(5) x root(5) = 3 root(5) / 5 That is, three times the square root of 5, divided by 5.
a+ square root of b has a conjugate a- square root of b and this is used rationalize the denominator when it contains a square root. If we want to multiply 5 x square root of 10 by something to get rid of the radical you can multiply it by square root of 10. But if we look at 5x( square root of 10 as ) 0+ 5x square root of 10 then the conjugate would be -5x square root of 10
[ 2 minus square root of 5 ] is the only one.
5
3 square root of 5 cannot be rationalised.3 square root of 5 cannot be rationalised.3 square root of 5 cannot be rationalised.3 square root of 5 cannot be rationalised.
Square root of 6
Square root (75) / square root (3) = 5
√3 = 1.73205081√5 = 2.23606798
17
You can get a decimal approximation with a calculator, with Excel, etc. But if you want to keep it as a square root, the "standard form" is considered to be one that has no square roots in the denominator. In this case, to get rid of the square root in the denominator, you multiply both numerator and denominator by the square root of 5, with the following result: 3 / root(5) = 3 root(5) / root(5) x root(5) = 3 root(5) / 5 That is, three times the square root of 5, divided by 5.
Square root 5: ± 2.236068Square root 9: ± 3
5 times the square root of two over 3 times the square root of two equals 5/3
I think it is the square root of 5.