Mathematics

Question

Does anyone know how to rationalise the denominator ???
Does anyone know how to rationalise the denominator ???

1 Answer

  • Answer:

    Multiply the entire expression by the radical denominator. This keeps the expression equal and makes the denominator 3, a whole number.

    Step-by-step explanation:

    The denominator is irrational because it is a radical number, √3.

    To rationalize the denominator, multiply the entire expression by √3. The denominator will become a whole number. This works because you multiply top and bottom by the same value.

    [tex]\frac{3 + \sqrt{2} }{\sqrt{3}}[/tex]

    [tex]= \frac{(\sqrt{3})(3 + \sqrt{2}) }{(\sqrt{3})(\sqrt{3})}[/tex]

    [tex]= \frac{(\sqrt{3})(3 + \sqrt{2}) }{3}[/tex]  

    √3 X √3 = 3 because:

    √3 X √3 = √3²

    Squaring a number and also finding its square root are opposites, or reverse operations. They cancel out

    You probably need to simplify the rest of the equation too.

    [tex]\frac{(\sqrt{3})(3 + \sqrt{2}) }{3}[/tex]

    [tex]= \frac{3(\sqrt{3}) + (\sqrt{2})(\sqrt{3})}{3}[/tex]  Distribute over brackets

    [tex]= \frac{3(\sqrt{3}) + \sqrt{2*3}}{3}[/tex]   Simplify

    [tex]= \frac{3(\sqrt{3}) + \sqrt{6}}{3}[/tex]

    Some people use a more simplified version whether they simplify each term in the numerator:

    [tex]\frac{3(\sqrt{3}) + \sqrt{6}}{3}[/tex]    (3√3)÷3 = √3

    [tex]\sqrt{3} + \frac{\sqrt{6}}{3}[/tex]

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