Question

if (x+2y) ² =x² + pxy +4y² , then the value of p is ________​

Answer 1

If x^2+4y^2=40 ……………(i) and xy=6

If xy=6 Therefore, x=6/y

Put x=6/y in (i)

(6/y)^2+4y^2=40

36/y^2+4y^2=40

(36+4y^4)/y^2=40

36+4y^4=40y^2 …………..(ii)

Let y^2=z

therefore, equation (ii) becomes

36+4z^2=40z ………. (iii)

Divide (iii) by 4 it becomes

9+z^2=10z

z^2–10z+9=0……………(iv)

z^2-z-9z+9=0

z(z-1)-9(z-1)=0

(z-1)(z-9)=0

Hence, Z=1 or 9

Taking Z=1, We have

y^2=1

y=1

Taking Z=9, we have

y^2=9

y=3

Putting y=1 in equation (i), we have

x^2+4(1)^2=40

x^2+4=40

x^2=36

x=6

Therefore, x+2y= 6+2(1)=6+2=8

Putting y=3 in equation (i) we have,

x^2+4(3)^2=40

x^2+4(9)=40

x^2=40–36

x^2=4

x=2

Therefore, x+2y= 2+2(3)=2+6=8

Answer 2

(x+2y)^2 = x^2+ pxy +4y^2
x^2 + 4xy + 4y^2 = x^2 + 4xy +4y^2
Then p = 4