Let b and c be constants such that the quadratic 2x^2+bx+c has roots 3±sqrt5 . Find the vertex of the graph of the equation y=-2x^2 + bx + c .
+128707 I'm assuming that we're talking about the quadratic 2x^2 + bx + c
If the roots are 3 ±sqrt5........then , by the quadratic formula -b/2a = 3 → -b = 2a* 3 = 2(2) * 3 = 12.......so.....b = -12
And since the irrational part of the root = sqrt(5).....we have that that this = 4sqrt(5)/[2a] =
4sqrt(5)/[2*2] = sqrt (16*5)/ 4 = sqrt (80) / 4
So.....this implies that b^2 - 4ac = 80 → 144 - 4(2)c = 80 → 64 = 8c → c = 8
So....our function is
y = 2x^2 -12x + 8
The x coordinate of the vertex = -b/2a = 12/ (2 *2) = 12/4 = 3
And we can find the y coordinate by subbing the previous result into the function....so we have
2(3)^2 - 12(3) + 8 = 18 - 36 + 8 = 26 - 36 = -10
So....the vertex is (3, -10)
P.S. - if the function is supposed to be -2x^2 + bx + c with the same roots, then b = 12 and c = -8 and IMAWESOME's answer is correct
Here's the graph of both functions : https://www.desmos.com/calculator/ewtfjezite
