Question

the arithmetic mean between a and b is twice the geometric mean between a and .prove that:7+4*1.73

Answer 1

Hey dear,

# Given-
A = 2G

# To prove-
a/b = 7 + 4√3

# Proof-
We have
A = 2G

Putting respective formulas,
(a + b) / 2 = √(ab)
(a + b) / (√ab) = 2
√(a/b) + √(b/a) = 2

Let a/b = x
√k + 1/√k = 2
Squaring both sides,
k + 1/k + 2 = 16
k^2 - 14k + 1 = 0

Solving this eqn,
k = 7 + 4√3 or k = 7 +- 4√3
a/b = 7 + 4√3 or a/b = 7 - 4√3

Hope this is useful...

Answer 2

Solution:

We know that arithmetic mean of two numbers a and b is given by

$= \frac{a + b}{2}$
Geometric mean of two numbers a and b is given by
$= \sqrt{ab}$
It is given that

Arithmetic mean = 2 Geometric mean

$\frac{a + b}{2} = 2 \sqrt{ab} \\ \\ a + b = 4 \sqrt{ab} \\ \\ \frac{a + b}{ \sqrt{a} \sqrt{b} } = 4 \\ \\ \frac{a}{ \sqrt{a} \sqrt{b} } + \frac{b}{ \sqrt{a} \sqrt{b} } = 4 \\ \\ \sqrt{ \frac{a}{b} } + \sqrt{ \frac{b}{a} } = 4 \\ \\ let \: \frac{a}{b} = u \\ \\ \sqrt{u} + \frac{1}{ \sqrt{u} } = 4 \\ \\ u + 1 = 4 \sqrt{u} \\ \\ {u}^{2} + 1 + 2u - 16u = 0 \\ \\ {u}^{2} - 14u + 1 = 0$
So apply Quadratic formula

$u_{1,2} = \frac{14 + - \sqrt{196 - 4} }{2} \\ \\ u_{1,2} = \frac{14 + - \sqrt{192} }{2} \\ \\ u_{1,2} = \frac{14 ± 8 \sqrt{3} }{2} \\ \\ u_{1} = 7 + 4 \sqrt{3} \\ \\ or \\ \\ u_{2} = 7 - 4 \sqrt{3} \\ \\ redo \: substitution \\ \\ \frac{a}{b} = 7 + 4 \sqrt{3} = 7+4*1.73\\ \\ \frac{a}{b} = 7 - 4 \sqrt{3} =7-4*1.73$

since √3 = 1.732