Question

compute the value of 9 X square + 4 y square if X Y is equal to 6 and 3 X + 2 Y is equal to 12​

Answer 1

Answer:

(a+b)^2 = a^2+ b^2+2ab

Therefore (3x+2y)^2 = 9x^2 + 4y^2 + 2xy .

= 16 + 12

= 28.

Step-by-step explanation:

Answer 2

$\sf\large\underline{Question:}$

Compute the value of 9x²+4y². If xy=6 and 3x+2y=12

$\sf\large\underline{Given:}$

• $\rm{xy=6------(i)}$
• $\rm{3x+2y=12-------(ii)}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\: value\:of\:9x^2+4y^2=?}$

$\sf\large\underline{Solution:}$

• [At 1st solve eq I and substituting Thier value in eq ii]

$\rm{\implies xy=6}$

$\rm{\implies x=\dfrac{6}{y}}$

• [Putting the value in eq ii here]

$\rm{\implies 3x+2y=12}$

$\rm{\implies 3(\dfrac{6}{y})+2y=12}$

$\rm{\implies \dfrac{18}{y}+2y=12}$

$\rm{\implies \dfrac{18+2y^2}{y}=12}$

$\rm{\implies 2y^2+18=12y}$

$\rm{\implies 2y^2-12y+18=0}$

• [Dividing by 2 on both sides]

$\rm{\implies y^2-6y+9=0}$

$\rm{\implies y^2-3y-3y+9=0}$

$\rm{\implies y(y-3)-3(y-3)=0}$

$\rm{\implies (y-3)(y-3)=0}$

$\rm{\implies \therefore y=3}$

• [Putting the value of y=3 in I]

$\rm{\implies xy=6}$

$\rm{\implies 3x=6}$

$\rm\red{\implies x=2}$

• [Now, calculate the 9x²+4y²=?]

$\rm{\implies 9x^2+4y^2}$

$\rm{\implies 9(2)^2+4(3)^2}$

$\rm{\implies 9*4+4*9}$

$\rm{\implies 36+36}$

$\rm{\implies 72}$

Hence,

$\rm\purple{\implies The\: value\:of\:9x^2+4y^2=72}$