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Zorluk: OrtaSurds and Rationalisation

If 6+26−2\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} is expressed in the form a+b3a + b\sqrt{3}, where aa and bb are integers, what is the value of a+ba + b?

  1. 3Cevap
  2. B
    2
  3. C
    5
  4. D
    1

Cevap

The value of a+ba + b is 3.
Multiplying both numerator and denominator by the conjugate (6+2)(\sqrt{6} + \sqrt{2}) yields 8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}. Comparing this to a+b3a + b\sqrt{3} gives a=2a = 2 and b=1b = 1, so a+b=3a + b = 3.

Adım Adım Çözüm

1
Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, (6+2)(\sqrt{6} + \sqrt{2}).
(6+2)(6+2)(6−2)(6+2)\frac{(\sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}
Multiplying by the conjugate eliminates radicals from the denominator using the difference of squares.
2
Expand the numerator (6+2)2(\sqrt{6} + \sqrt{2})^2 using (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2.
(6)2+212+(2)2=6+2(23)+2=8+43(\sqrt{6})^2 + 2\sqrt{12} + (\sqrt{2})^2 = 6 + 2(2\sqrt{3}) + 2 = 8 + 4\sqrt{3}
Simplifying 12=4×3=23\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} converts compound surds to standard form.
3
Expand the denominator using the difference of two squares (x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2.
(6)2−(2)2=6−2=4(\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4
Squaring each square root leaves rational integers in the denominator.
4
Divide the expanded numerator by the denominator to find aa and bb.
8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}
Dividing each term by 4 gives a=2a = 2 and b=1b = 1.
5
Compute the sum a+ba + b.
2+1=32 + 1 = 3
Adding the coefficients aa and bb yields the required value.

Anahtar Kavram

Rationalisation of Binomial Denominators
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Konular

Surds and Rationalisation

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