Question

a+b=3 ab=2 then a^4+b^4​

Answer 1

Answer:

Given ,

• a + b = 3
• ab = 2

To find :-

• a⁴ + b⁴

Now,

a⁴ + b⁴

=( a²)² + (b²)²

= ( a²)² + (b²)² + 2a²b² - 2a²b²

= (a²+b²)² - 2a²b²

= ( a² + b² + 2ab - 2ab ) - 2a²b²

= (a+b)²-2ab - 2a²b²

= (a+b)² - 2ab - 2(ab)²

Substituting the values ,

3² - 2× 2 - 2×2²

= 9 - 4 - 8

= 9 - 12

= $\boxed{\bold{-3}}$

Answer 2

Answer:

Step-by-step explanation:

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

because

w.k.t  $(a^2 + b^2)^2$ = $a^4 + b^4 + 2a^2b^2$

subtracting $2a^2b^2$ on both side,

we get $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$

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

Similarly,

by applying the same logic to $(a^2 + b^2)$

we finally end up with this expression,

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

Now by substituting and calculating carefully,

th solution is....

$a^4 + b^4 =$ 17

or

simply by seeing the the first 2 eqs, we can deduce that a and b are 1 and 3.