Knowledge Check (Quadratic Equation)

Questions

1. Express in terms of k and h, the sum of root and product of roots of x² + kx +h.

2. Given 2x² + kx +8 is always positive, find the range of values of k.

3.  Complete the square for 2x² + 3x -4, express it in the form of a(x-b)² + c.

4. Sketch the curve y= x² + 3x -4, -5 ≤x≤ 3.

5. Find the range of values of 9 - x² > 0.

6. Find the range of values of x² + 5x - 6 < 0.

7. Given y =2x² + kx+ 8 intersects y= x at 2 distinct points. Find the range of values of k.

8. Given  y= x is tangent to y =2x² + kx+ 8, find the values of k.

9. Given  y= x is does not intersect  y =2x² + kx+ 8, find the range of values of k.

Answers

1	Sum of roots = -k; Product of roots = h.
2	Discriminant < 0 k2 - 4(8)(2)< 0 k2 - 64 < 0  (k + 8)(k-8) <0 -8 < k < 8
3	a = 2; b=3/4 ; c= -71/16
4	y= x2 + 3x -4, -5 ≤x≤ 3   = (x+4)(x-1) x -5 -4 -1.5 0 1 3 y 6 0 -6.25 -4 0 14
5	9 - x2 > 0 x2 -9 > 0  (x+ 3)(x-3) > 0 x> 3 or x< -3
6	x2 + 5x - 6<0 (x +6)(x-1) <0 -6 < x <1
7	y =2x2 + kx+ 8 ----- (1) y= x                 ------(2) 2x2 + kx+ 8 = x 2x2 +( k-1)x+ 8 =0 Apply discriminant > 0 ( k-1)2 - 4(2)(8)>0  k2 -2k+1-64 >0 k2 -2k-63 >0 (k-9)(k+7)>0 k> 9 or k<-7
8	y =2x2 + kx+ 8 ----- (1) y= x                 ------(2) 2x2 + kx+ 8 = x 2x2 +( k-1)x+ 8 =0 Apply discriminant =  0 ( k-1)2 - 4(2)(8)= 0  k2 -2k+1-64 = 0 k2 -2k-63 = 0 (k-9)(k+7)= 0 k= -7 or k = 9
9	y =2x2 + kx+ 8 ----- (1) y= x                 ------(2) 2x2 + kx+ 8 = x 2x2 +( k-1)x+ 8 =0 Apply discriminant <  0 ( k-1)2 - 4(2)(8)< 0  k2 -2k+1-64 < 0 k2 -2k-63 < 0 (k-9)(k+7)< 0 -7 <k< 9