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Worked Examples · Example 28

Q.If the AM and GM of two positive numbers xx and yy are 13 and 12 respectively, find the numbers.

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Use the AM and GM definitions to get x+yx+y and xyxy, then solve the resulting quadratic.

[!FORMULA] For two positive numbers x,yx,y: AM=x+y2\text{AM}=\dfrac{x+y}{2}, GM=xy\text{GM}=\sqrt{xy}. They are the roots of t2−(x+y)t+xy=0t^2-(x+y)t+xy=0.

  1. AM=13 ⇒ x+y2=13 ⇒ x+y=26\text{AM}=13\ \Rightarrow\ \dfrac{x+y}{2}=13\ \Rightarrow\ x+y=26.
  2. GM=12 ⇒ xy=12 ⇒ xy=144\text{GM}=12\ \Rightarrow\ \sqrt{xy}=12\ \Rightarrow\ xy=144.
  3. x,yx,y are roots of t2−26t+144=0t^2-26t+144=0.
  4. Discriminant: D=262−4(144)=676−576=100D=26^2-4(144)=676-576=100, D=10\sqrt{D}=10.
  5. t=26±102t=\dfrac{26\pm10}{2}, giving t=18t=18 or t=8t=8.
  6. Check: 18+8=2618+8=26 ✓, 18×8=144=12\sqrt{18\times8}=\sqrt{144}=12 ✓.
✓Final answer

The numbers are 1818 and 88

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