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Exercise 5.2 · Q10

Q.If the AM of two unequal positive real numbers aa and bb (a>ba > b) is twice as much as their G.M., show that a:b=(2+3):(2−3)a : b = (2+\sqrt{3}) : (2-\sqrt{3}).

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Convert the AM =2×=2\timesGM condition into a quadratic in a/b\sqrt{a/b} and solve.

AM=a+b2\text{AM}=\dfrac{a+b}{2}, GM=ab\text{GM}=\sqrt{ab}.

  1. Given a+b2=2ab⇒a+b=4ab\dfrac{a+b}{2}=2\sqrt{ab}\Rightarrow a+b=4\sqrt{ab}.
  2. Let k=abk=\dfrac{a}{b} (so k>1k>1 since a>ba>b) and x=kx=\sqrt{k}.
  3. Divide the equation by bb: ab+1=4ab⇒k+1=4x⇒x2+1=4x\dfrac{a}{b}+1=4\sqrt{\dfrac{a}{b}}\Rightarrow k+1=4x\Rightarrow x^2+1=4x.
  4. Rearranged: x2−4x+1=0⇒x=4±16−42=4±232=2±3x^2-4x+1=0\Rightarrow x=\dfrac{4\pm\sqrt{16-4}}{2}=\dfrac{4\pm2\sqrt3}{2}=2\pm\sqrt3.
  5. Since a>ba>b, x=a/b>1x=\sqrt{a/b}>1; as 2−3≈0.27<12-\sqrt3\approx0.27<1, reject it, so x=2+3x=2+\sqrt3. …

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