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If x and y are two positive real numbers such that x 2 +4y 2 =17 and xy =2, then find the value of x- 2y. a. 3 b. 4 c. 8 d. 9

Answer :

JeanaShupp

Answer: The value of x- 2y is a. [tex]\pm 3[/tex].

Step-by-step explanation:

Given:  x and y are two positive real numbers such that [tex]x^2+4y^2=17[/tex]   and [tex]xy= 2[/tex] .

Consider [tex](x-2y)^2=x^2-2(x)(2y)+(2y)^2\ \ \ [(a+b)^2=(a^2-2ab+b^2)][/tex]

[tex]=x^2-4xy+4y^2[/tex]

[tex]=x^2+4y^2-4(xy)[/tex]

Put  [tex]x^2+4y^2=17[/tex]   and [tex]xy= 2[/tex] , we get

[tex](x-2y)^2=17-4(2)=17-8=9[/tex]

[tex]\Rightarrow\ (x-2y)^2=9[/tex]

Taking square root on both sides , we get'

[tex]x-2y= \pm3[/tex]

Hence, the value of x- 2y is a. [tex]\pm 3[/tex].

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