Question

If a^2+b^2=58 and a-b=4 then the value of ab is

Answer 1

Answer:

21

Step-by-step explanation:

As we know that (A-b)²=a²+b²-2ab

=(a-b)²+2ab=a²+b² [2ab move to LHS and become positive]

=(4)²+2ab=58

=16+2ab=58

=2ab=58-16 [16 moves to RHS and become negative]

=ab=42/2 [×2 moves to RHS and become divident]

=ab=21

Answer 2

Given: a^2+b^2=58 and a-b=4

To find: The value of ab.

Solution:

When the term (a-b) is squared, the formula of it can be written as follows.

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

Now, the values of the term (a²+b²) and the term (a-b) are substituted in the formula as given in the question. This will leave (ab) as the only variable in the formula.

$(4)^{2} = 58 - ab$

Now, the value of ab is found by rearranging the formula and calculating for ab.

$ab = 42$

Therefore, the value of ab is 42.