Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate x^2 + 2x + 4 to zero:

x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:

rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))

or x = -1 plus or minus i*sqrt(3)

This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.

3 0

3 years ago

The point (4,-2) is the vertex of the graph of a quadratic function. The points (8,6) and (2,0) also fall on the graph of the fu

Elina [12.6K]

Answer:

Step-by-step explanation:

y = ax² + bx + c

~~~~~~~

(4, - 2), (8, 6), (2, 0)

a(4²) + b(4) + c = - 2

a(8²) + b(8) + c = 6

a(2²) + b(2) + c = 0

16a + 4b + c = - 2 .............. (1)

64a + 8b + c = 6 ............... (2)

4a + 2b + c = 0 ................. (3)

a = 0.5 ; b = - 4 ; c = 6

y = 0.5x² - 4x + 6

Part A: (0, 6), (6, 0)

Part B: (0,6)

Part C: (6, 0)

Part D: [ 4, ∞ )

4 0

2 years ago

Solve each question using the quadratic formula? X2+4x+3=0

Sergeeva-Olga [200]

x=−1 or x=−3 i think lol

8 0

3 years ago