Question

if x and y are two positive real numbers such that x2+4y2=17 and xy=2 then find the value of (x+2y)

Answer 1

hope it will help u ................

Answer 2

Concept:

As per algebraic identities we know that, the square of sum of two number is given by the sum of the sum of squares of that two numbers and twice of product of that numbers.

In mathematical words we can say, $(a+b)^2=a^2+b^2+2ab$

Given:

Given that, x and y are two positive real numbers such that $x^2+4y^2=17$ and xy=2.

Find:

The value of (x+2y).

Solution:

Given that,  $x^2+4y^2=17$ and xy=2.

So now calculating we get,

$(x+2y)^2$

$=x^2+(2y)^2+2.x.2y$, using the algebraic identity

$=x^2+4y^2+4xy$

= 17 + 4*2 , substituting the values of the given relations

= 17 + 8

= 25

So, now the value is $(x+2y)^2=25$

Therefore, x+2y = $\sqrt{25}=\pm5$

Since x and y are positive so the sum of positive cannot be negative so we ignore the negative value of x+2y.

Hence the value of x + 2y = 5.

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