Question

if 4x square +y square = 40 and xy =6 so find the value of 2x +y​

Answer 1

$\huge\mathbb{SOLUTION:-}$

Given :-

$\sf \implies 4x{}^{2}+y{}^{2}=40\\ \sf\implies xy=60$

• we have to find the value of 2x+y

$\sf Formula\:will\:be\:used:-\\ \\ \sf(a+b){}^{2}=a{}^{2}+b{}^{2}+2ab$

$\bf\underline{According\:to\: Question}$

$\sf (2x+y){}^{2}=(2x){}^{2}+y{}^{2}+2\times 2x\times y\\ \\ \sf\rightarrow (2x+y){}^{2}=4x{}^{2}+y{}^{2}+8xy\\ \\ \tt\implies we\:have=2x+y=40\: \: \: \: \: xy=6\\ \\ \sf\rightarrow (2x+y){}^{2}=40+4\times 6\\ \\ \sf\rightarrow (2x+y){}^{2}=40+24\\ \\ \sf\rightarrow 2x+y=\sqrt{64}\\ \\ \tt\implies 2x+y=8$

•The value of 2x+y is 8

#BAL

#answerwithquality

Answer 2

$\underline \mathbb{SOLUTION}$

$\underline \mathbb{GIVEN}$

$= > 4 {x}^{2} + {y}^{2} = 40$

$= > xy = 6$

NOW,

TO FIND THE VALUE OF 2x + y

$= > identity = (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\underline \mathbb{ATQ}$

$= > {(2x + y)}^{2} = {(2x)}^{2} + {y}^{2} + 2(2x)(y)$

$= > {(2x + y)}^{2} = 4 {x}^{2} + {y}^{2} + 4xy$

WE HAVE,

$= > 2x + y = 40 \: and \: xy = 6$

$= > {(2x + y)}^{2} = 40 + 4 \times 6$

$= > {(2x + y)}^{2} = 40 + 24$

$= > 2x + y = \sqrt{64}$

$= > 2x + y = 8$

ANSWER = 2x + y = 8

_______________________________________

#BAL

#Answerwithquality