Step-by-step explanation:
For quadratic equation ax^2 + bx + c = 0 to have two distinct real roots,
b^2 - 4ac must be positive.
b^2 - 4ac > 0
(k - 3)^2 - 4(3 - 2k) > 0
k^2 - 6k + 9 - 12 + 8k > 0
k^2 + 2k - 3 > 0
Answer:
(- 1, 1) and (- 5, - 35)
Step-by-step explanation:
since both equations express y in terms of x, equate the right sides
x² + 15x + 15 = - x² + 3x + 5 ( subtract - x² + 3x + 5 from both sides )
2x² + 12x + 10 = 0 ( divide all terms by 2 )
x² + 6x + 5 = 0 ← in standard form
(x + 1)(x + 5) = 0 ← in factored form
equate each factor to zero and solve for x
x + 1 = 0 ⇒ x = - 1
x + 5 = 0 ⇒ x = - 5
substitute these values into either of the 2 equations and solve for y
x = - 1 : y = (- 1)² + 15(- 1) + 15 = 1 - 15 + 15 = 1 ⇒ (- 1, 1 )
x = - 5 : y = (- 5)² + 15(- 5) + 15 = 25 - 75 + 15 = - 35 ⇒ (- 5, - 35 )
None of the given options
