Question

if 3x²+3y²=42 and xy=15.find the value of (x-y)²​

Answer 1

Step-by-step explanation:

Given:-

$\\ \tt{:}\rightarrowtail 3x^2+3y^2=42$

$\\ \tt{:}\rightarrowtail xy=15$

To find:-

$\\ \tt{:}\rightarrowtail (x-y)^2$

Solution:-

$\\ \tt{:}\rightarrowtail 3x^2+3y^2=42$

$\\ \tt{:}\rightarrowtail 3(x^2+y^2)=42$

$\\ \tt{:}\rightarrowtail x^2+y^2=\dfrac{42}{3}$

$\\ \tt{:}\rightarrowtail x^2+y^2=14$

we know that

$\boxed{\sf (x-y)^2=x^2+y^2-2xy}$

• Substitute the values

$\\ \tt{:}\rightarrowtail 14-2(15)$

$\\ \tt{:}\rightarrowtail 14-30$

$\\ \tt{:}\rightarrowtail (x-y)^2=-16$