Question

A, b and c are positive integer such that a^2+b^2-2bc=100 and 2ab-c^2=100 then the value of (a+b) /c is

Answer 1

Correct question should be:

A, b and c are positive integer such that a² + 2b²- 2bc = 100 and 2ab - c² = 100 then the value of (a+b) /c is?

Given:

a, b & c are positive integers such that

a² + 2b² - 2bc = 100 ........ (i)

and

2ab - c² = 100 ....... (ii)

To find:

$\frac{a\: +\: b}{c}$

Solution:

Now,

On comparing eq. (i) & (ii), we get

a² + 2b² - 2bc = 2ab - c²

⇒ a² + b² + b² - 2bc = 2ab - c²

⇒ a² + b² + b² - 2bc - 2ab + c² = 0

⇒ [a² - 2ab + b²] + [b² - 2bc + c²] = 0

We know: (a - b)² = a² - 2ab + b²

⇒ [a - b]² + [b - c]² = 0

∴ [a - b]² = 0 & [b - c]² = 0

⇒ [a - b] = 0 & [b - c] = 0

⇒ a = b & b = c

⇒ a = b = c

Let's assume a = b = c = "x"

We will substitute a = b = c = x in eq. (i)

x² + 2x² - 2xx = 100

⇒ x² + 2x² - 2x² = 100

⇒ x² = 100

⇒ x = ± 10

We will now substitute the value a = b = c = + 10 in (a+b)/c, we get

= $\frac{10 + 10}{10}$

= $\frac{20}{10}$

= 2

Similarly, substituting a = b = c = - 10 in (a+b)/c, we get

= $\frac{-10 - 10}{-10}$

= $\frac{-20}{-10}$

= 2

Thus, $\boxed{The \:value\: of \:\frac{a\:+\:b}{c} \:is\:\bold{ 2}}.$

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