Determine the equation of a quadratic in standard form if it has x-intercepts of 1 −√2 and 1 + √2, and it passes through the point (2,4)

Question

1. Determine the equation of a quadratic in standard form if it has x-intercepts of 1 −√2 and 1 + √2, and it passes through the point (2,4)
2. For what value of k does the equation kx−8=2x² have one real root?
3. The ball is kicked up in the air and lands down the field, according to the following equation: h = −5t + 20t, where h = height in meters and t= time in seconds.
4. a) how long is the ball above 3 meters?
5. b) what is the maximum height of the football and at what time does the maximum height occur?

Answer 1

x = 1 + sqr2

x-1 = sqr2

(x-1)^2 = (sqr2)^2

x^2 -2x + 1 = 2

x^2 -2x - 1 = 0

y = a(x^2 -2x -1)

4 = a((2)^2 -2(2) -1)

4 = a(4-4-1)

4 = -a

a=-4

f(x) = -4(x^2 -2x -1)

f(x) = -4x^2 +8x +4 is the quadratic with the given zeros through (2,4)

kx-8 = 2x^2

has one real root when the discriminant = 0,

b^2 -4ac = 0,

k^2 -4(2)(8)=0,

k^2-64 =0

k =+8 or -8

kx -8 = 2x^2

8x-8 = 2x^2

2x^2 -8x +8 = 0

x^2 -4x+4 = 0

(x-2)^2 = 0

x = 2 is the only real solution when k=8

2x^2 +8x +8 = 0

x^2 +4x +4 = 0

(x+2)^2 = 0

x =-2 is the only real solution when k=-8

h(t) = -5t ^2+ 20t

h'(t) =-10t +20 = 0

10t = 20

t =20/2

t = 2 seconds when the ball is at maximum height

h(2) = -5(2)^2 +20(2) = 20 meters = maximum height

3 = -5t^2 + 20t

5t^2 -20t +3 = 0

t = 20/10 + or - (1/10)sqr(20^2 -4(5)(3))

t = 2 + or -(1/10)sqr340

t = 2+1.844 or 2-1.844

t = 0.156 seconds to 3.844 seconds the balll is above 3 meters

3.844-0.156

= 3.688 seconds the ball is above 3 meter