Question

given that a+b=2 and that a^2+b^2=6 prove that 2ab=-2.
find also the value of (a-b)^2

Answer 1

(a-b)^2 =(a+b)^2-4ab

2^2-2(2ab)

0

Answer 2

The correct answer is if a + b=2 and $a^{2} +b^{2} =6$ then 2ab=-2.

Given: The equation = a + b=2 and $a^{2} +b^{2} =6$.

To Prove: 2ab = -2.

Solution:

The equation = a + b=2 and $a^{2} +b^{2} =6$.

Squaring both the the sides in the equation a + b = 2.

$(a+b)^{2} =2^{2}$

$a^{2} +2ab+b^{2} =4$

Put the value of $a^{2} +b^{2} =6$ in this equation.

6 + 2ab = 4

2ab = -2

Hence proved.

#SPJ2