Question

. If a +b = 3 and a - b = 2, what is the value of
8ab(a2 + b2)?​

Answer 1

Answer:-

Given:

a + b = 3 -- equation (1)

a - b = 2 -- equation (2)

Add equations (1) & (2).

⟹ a + b + a - b = 3 + 2

⟹ 2a = 5

⟹ a = 5/2

Substitute the value of a in equation (1).

⟹ 5/2 - b = 3

⟹ 5/2 - 3 = b

⟹ (5 - 6) / 2 = b

⟹ - 1/2 = b

We have to find:

⟹ 8ab (a² + b²)

Putting the values we get,

⟹ 8(5/2)( - 1/2) [ (5/2)² + ( - 1/2)² ]

⟹ ( - 10) [ 25/4 + 1/4 ]

⟹ ( - 10) [ (25 + 1) / 4 ]

⟹ ( - 10) ( 26/4)

⟹ - 65

∴ The required answer is - 65.

Answer 2

Answer:

${ \huge{ \underline{ \large{ \rm { \pink{Given:}}}}}}$

• a + b = 3.......(1)
• a - b = 2 ...... (2)

$\huge{ \underline{ \rm{ \large{ \red{Find:}}}}}$

• 8ab(a² + b²)

$\huge{ \underline{ \rm{ \large{ \green{Solution:}}}}}$

Add equation (1) & (2)

⇒a + b + a - b = 3 + 2

⇒ 2a = 5

⇒ a = 5/2

⠀⠀⠀⠀⠀⠀⠀⠀⠀

So, value of a is 5/2 .

⠀

Let us substitute the value of a in equation(1)

⇒ a + b = 3

⇒ 5/2 + b = 3

⇒ b = 3 - 5/2

⇒ b = -1/2

⠀

So, Value of b is -1/2

Now let us find 8ab (a² + b²) ,

⠀

${ \to{8 \times \frac{5}{2} \times \frac{ - 1}{2} (( { \frac{5}{2}) }^{2} + { (\frac{ - 1}{2}) }^{2}) }}$

${ \to{4 \times 5 \times \frac{ - 1}{2} ( \frac{25}{4} + \frac{1}{4}) }}$

${ \to{20 \times \frac{ - 1}{2}( \frac{25 + 1}{4} ) }}$

${ \to{ \frac{ - 20}{2} \times \frac{26}{4} }}$

${ \to{ - 5 \times 13 = - 65 }}$

${ \therefore{ \sf{ \blue{Value \: of \: 8ab(a² + b²) \: is \: -65}}}}$