Positive Roots of Quadratic Equations
Yue Kwok Choy
Question
Given that the quadratic equation
find the range of k .
x2 + (k + 2)x + (k + 5) = 0 has two positive roots,
"Solution"
Let ,  be the roots of the quadratic equation x2 + (k + 2)x + (k + 5) = 0 .
Since ,  are both positive, therefore  +  > 0 and  > 0 .
By Vieta's Theorem, we have :

    (k  2)  0 k  2


   k  5  0
k  5
The range of k is
–5 < k < –2 .
Analysis
" +  > 0 and  > 0" does not imply that the quadratic equation has positive roots.
The reason is that the quadratic equation may have complex roots (not real roots). In this case, the
sum and product of roots can still be positive. Take an example, if k = –3 which is within the
range, the quadratic equation becomes x2 – x + 2 = 0 .
The discriminant
x
 = (–1)2 – 4(1)(2) = –7 < 0, the equation has complex roots:
1  7i
2
Solution
Let ,  be the roots of the quadratic equation
By Vieta's Theorem, we have :
x2 + (k + 2)x + (k + 5) = 0 .
  k  22  4k  5  0 k  4k  4  0 k  4 or k  4



k  2

k  2
     (k  2)  0  



  k  5  0
k  5
k  5




The range of k is
–5 < k < –4 .
Lessons
Given ,  be the roots of the quadratic equation :
ax2 + bx + c = 0 (*) , a  0, a, b, c are real numbers.
1.

  b 2  4ac  0

b
(*) has positive roots         0
a

c
    0

a
2.

  b 2  4ac  0

b
(*) has negative roots         0
a

c
    0

a
3.
(*) has exactly one zero root

b  0

c  0
4.
(*) has two zero roots

b  0

c  0
5.
(*) has one positive root and one negative root
  b 2  4ac  0
c
 
   0

a