Question

if a+b=10 and a-b=6 then find the values of ab and a²+b²​

Answer 1

Answer:

68

Step-by-step explanation

Give that

           a+b=10 ...........  (1)

           a-b=6 ................. (2)

from equation (1)  we get

                                       a=6+b { substute this in the equation (1) }

                              we get

                                          (6+b)+b = 10

                                           6+2b = 10

                                           2b = 10-6 = 4

                                            b = 4/2 = 2

                            therefore b = 2   { substute value of b in equation (2) }

                             we get

                                          a-2 = 6

                                          a = 6+2

                             therefore a = 8

Now  

       ab = (8)(2)

            =16

      a²+b² = 8²+2²

                =64+4

                =68

Answer 2

Answer:

Required value of ab is 16 and value of a²+b² is 68.

Step-by-step explanation:

Given,

$a + b = 10$

and

$a - b = 6$

Here we want to find

${a}^{2} + {b}^{2}$

We know,

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

Now,

$a - b = 6$

We are squaring both sides,

${(a - b)}^{2} = {6}^{2} \\ {(a - b)}^{2 } = 36 \\ (a + b)^{2} - 4ab = 36 \\ {10}^{2} - 4ab = 36 \\ 100 - 4ab = 36 \\ 4ab = 100 - 36 \\ 4ab = 64 \\ ab = \frac{64}{4} \\ ab = 16$

So,

${a}^{2} + {b}^{2} \\ = {(a + b)}^{2} - 2ab \\ = {10}^{2} - 2 \times 16 \\ = 100 - 32 \\ = 68$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549