Find two positive numbers a and b, whose AM 10 and GM 8

\[AM = 10\]

\[ \therefore \frac{a + b}{2} = 10\]

\[ \Rightarrow a + b = 20 . . . . . . . . (i)\]

\[\text { Also }, G = 8\]

\[ \therefore \sqrt{ab} = 8\]

\[ \Rightarrow ab = 8^2 \]

\[ \Rightarrow ab = 64 . . . . . . . . (ii)\]

\[\text { Using (i) and (ii) }: \]

\[ \Rightarrow a\left( 20 - a \right) = 64\]

\[ \Rightarrow a^2 - 20a + 64 = 0\]

\[ \Rightarrow a^2 - 16a - 4a + 64 = 0\]

\[ \Rightarrow a\left( a - 16 \right) - 4\left( a - 16 \right) = 0\]

\[ \Rightarrow \left( a - 16 \right)\left( a - 4 \right) = 0\]

\[ \Rightarrow a = 4, 16\]

\[\text { If a = 4, then b = 16 } . \]

\[\text { And, if a = 16, then b = 4 .} \]

a+b/2 = 34 hence a+b = 68 .256 = ab.from here, u get : a(68-a)=256a2-68a+256=0(a-4)(a-64)=0hence,a= 4   or a = 64

b=64  or b = 4

hence numbers are 4 and 64.

Text Solution

Solution : we have, <br> `(a+b)/2=10` and `sqrt(ab)=8=>a+b=20` and `ab=64` <br> Clearly,`a` and `b` are roots of the equation <br> `x^2-20x+64=0` <br> `(x-16)(x-4)=0=>x=4,16=>a=4,b=16=>a=16,b=4`

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