Question

If 2x^2 - (2 + k)x + k = 0 where k is a real number,
find the roots of the equation.​

Answer 1

Answer:

ROOT FORMULA

= {-b±√(b²-4ac)}/2a

now

= [2+k±√{(2+k)²-8k}]/4

= [2+k ±√(4+k²+4k-8k)]/4

First root is (2+k)/4+√(k²-4k+4)/4

Second root is (2+k)/4-√(k²-4k+4)/4

Answer 2

Answer:

Step-by-step explanation:

If k is a real number then the roots of the quation must be real also

So, D =》 b² - 4ac =》 (2 + k)² - 4×2×k

=》4 + 4k + k² - 8k =》 4 - 4k + k²

=》 (2 - k)²

Roots of the equation =

(-b+-√D)/2a =》 {-(2 + k)+-√(2 - k)²}/2 × 2

=》 (- 2 - k)+-(2-k)/4

=》 (- 2 - k + 2 - k)/4 , (- 2 - k - 2 + k)/4

=》 - 2k/4 , - 4/4

=》 - k/2 , - 1

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