A quadratic of the form $-2x^2 + bx + c$ has roots of $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.$ The graph of $y = -2x^2 + bx + c$ is a parabola. Find the vertex of this parabola. Help needed ASAP

Guest Jul 18, 2020

#2**0 **

y = -2x^{2} + bx + c with roots 3 + sqrt(5) and 3 - sqrt(5)

One answer is: y = [ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ]

Let's multiply out the right-hand side:

[ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ] ---> [ x - 3 - sqrt(5) ] · [ x - 3 + sqrt(5) ]

---> x^{2} - 3x + sqrt(5)·x - 3x + 9 - 3·sqrt(5) - sqrt(5)·x + 3·sqrt(5) - 5

---> x^{2} - 6x + 4

However, since the original problem started with "-2x^{2}", we'll have to multiply this answer by -2:

---> y = -2x^{2} + 12x - 8

To find the vertex: y = -2x^{2} + 12x - 8

---> add 8 to both sides: y + 8 = -2x^{2} + 12x

---> factor out the -2: y + 8 = -2(x^{2} - 6x)

---> complete the square: y + 8 - 18 = -2(x^{2} - 6x + 9)

y - 10 = -2(x - 3)^{2}

The vertex is (3, 10)

geno3141 Jul 18, 2020