The line joining the vertex of fx x2kx 14 and 00 has slope

The line joining the vertex of f(x) = -x^2-kx + 1/4 and (0.0) has slope equals 1 if k equals.... A polynomial of degree 5 with zeros i-1.-1, and -Squareroot 5. must have a constant term divisable by.... The complex number 3 + i/2i simplifies to: f(x) = (5x^2 + 4)(3x^2 + 2)(x + 1) has the zeros:

Solution

1. The vertex of f(x) = -x^2 -kx + 1/4

x = -b/2a = -(-k)/(2*-1) = -k/2

f(x) = -k^2/4 -k(-k/2) +1/4 = -k^2/4 +k^2/2 +1/4 = ( -k^2 +2k^2+1)/4 = (k^2 +1)/4

We have one point ( x, f(x) and another point ( 0,0)

Find the slope from these two points:

m = 1 ; {(k^2 +1)/4 -0}/( -k/2) = (k^2 +1)/-2k

(k^2 +1)/-2k = 1

k^2 + 1 +2k =0

(k+1)^2 = 0 ------> k = -1

3. ( 3+i)/2i

Rationalise both numerator and denominator

(3+i)*2i/(2i*2i)

= (6i - 4)/(-4)

= - 3i/2 + 16

4. f(x) = (5x^2 +4)( 3x^2 + 2)(x +1)

To find zeros : f(x) =0;

(5x^2 +4)( 3x^2 + 2)(x +1) =0

x = -1 ; 5x^2 =- 4

x = +/- 2i/sqrt5

3x^2 +2 =0

x = +/- isqrt(2/3)