Question

If x^2+y^2=26and x =3 , find the value of a) x+y b) x-y c) x^4 + y^4​

Answer 1

Answer is given in the attachment. Hope it is clear. please do cross check my answer.

Answer 2

Answer:

a)3 + √17

b)3 - √17

c)370

Step-by-step explanation:

Given,

x² + y² = 26----1

x = 3

x² = 9

Putting 2 in one we get,

3² + y² = 26

y² = 26 - 9 = 17----3

y = √17

a) x + y = 3 + √17

b) x - y = 3 - √17

c)Here we can do in 2 methods

1. We know that,

x² + y² = 26

Squaring both sides we get,

(x² + y²)² = 26²

(x²)² + 2x²y² + (y²)² = 676

x⁴ + 2x²y² + y⁴ = 676

x⁴ + y⁴ = 676 - 2x²y²

From 2 and 3 we get,

x⁴ + y⁴ = 676 -2(9)(17)

x⁴ + y⁴ = 676 - 306 = 370

OR

x = 3 and y = √17

x⁴ = 3⁴ = 3 × 3 × 3 × 3 = 9 × 9 = 81----4

y⁴ = (√17)⁴ = (√17) × (√17) × (√17) × (√17)

If a square rooted number is square it is the number itself

let's check √25 = 5

squaring both sides we get

(√25)² = 5² = 25

so (√25)² = (√25) × (√25) = 25

Now, y⁴ = (√17) × (√17) × (√17) × (√17) = 17 × 17 = 289----5

so from 4 and 5 we get

x⁴ + y⁴ = 81 + 289 = 370

Hope you understood it........All the best