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If x^(2)+y^(2)=k^(2), and xy=8-4k, what ...

If `x^(2)+y^(2)=k^(2)`, and `xy=8-4k`, what is `(x+y)^(2)` in terms of k ?

A

`k-4`

B

`(k-4)^(2)`

C

`k^(2)-4k+8`

D

`(k-2)^(2)+4`

Text Solution

AI Generated Solution

The correct Answer is:
To find \((x+y)^2\) in terms of \(k\), we start with the given equations: 1. \(x^2 + y^2 = k^2\) 2. \(xy = 8 - 4k\) We know from algebra that: \[ (x+y)^2 = x^2 + y^2 + 2xy \] Now, we can substitute the values from the given equations into this identity. ### Step 1: Substitute \(x^2 + y^2\) and \(xy\) Substituting the values we have: \[ (x+y)^2 = (x^2 + y^2) + 2(xy) \] Substituting \(x^2 + y^2 = k^2\) and \(xy = 8 - 4k\): \[ (x+y)^2 = k^2 + 2(8 - 4k) \] ### Step 2: Simplify the expression Now, we simplify the right-hand side: \[ (x+y)^2 = k^2 + 16 - 8k \] ### Step 3: Rearranging the terms Rearranging the terms gives us: \[ (x+y)^2 = k^2 - 8k + 16 \] ### Step 4: Recognize the perfect square Notice that \(k^2 - 8k + 16\) can be expressed as a perfect square: \[ (x+y)^2 = (k - 4)^2 \] ### Conclusion Thus, we find that: \[ (x+y)^2 = (k - 4)^2 \] ### Final Answer The value of \((x+y)^2\) in terms of \(k\) is \((k - 4)^2\). ---
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