Download Name

M$5: Intermediate Algebra & Trigonometry
Mr. Tamhane
2009-2010
Name: _______________________
Date: _______________________
Sum and Product of the Roots of Quadratic Equations Worksheet
1. One root of x 2  kx  12  0 is 3. Find k.
Product = 12, so 3(x) = 12
x=4

3 +4 = 7, but I’ll use -7 in the equation. So, k=-7
2. The roots of the equation x 2  4 x  c  0 are 2+3i and 2-3i. What is the value of c?
C is the product of the two roots.
(2+3i)(2-3i)=4-6i+6i-9i2 = 4 + 9 = 13

3. A quadratic equation having the roots 0 and -2 is:
Sum: -2, but I’ll use 2 in the equation
Product: 0
So my answer is x2+2x+0=0 or more simply x2+2x=0
4. Write, in simplest form, a quadratic equation whose roots are k and –k.
Sum: 0
Product: -k2
So x2+0x-k2=0 or more simply, x2-k2=0
5. Given the equation x 2  kx  7  0 , and knowing that the discriminant is greater than
zero and a perfect square:
a. Find k if k is negative.
b. Find k if k is positive.

The roots are real, rational, and unequal. So, we need two numbers that multiply to 7 that fit that
criteria. There are multiple answers, but two options for the roots might be: {-7, -1} or {1, 7}
-1 + -7 = -8, so k  8 (the answer to b)
1 + 7 = 8 so k  -8 (the answer to a)
6. For the following equation x 2  bx  c  0 , if c<0, there are two possibilities for what
the roots may be like. Describe these two possibilities and what would have to be
true for each situation to occur. Hint: think about the discriminant.

If c<0, then b2 – 4ac must be greater than 0 (since a is 1, 4ac will be negative, and if you subtract
a negative number, you add the positive number).
Then if the discriminant is a perfect square, the roots will be real, rational, and distinct
If the discriminant is not a perfect square, the roots will be real, irrational, and distinct
5
7. One root of the equation 3x 2  x  k  0 is  . Find the other root of the equation.
3
As the equation is written now, we do not know the sum of the roots, since a =/=1. We can
divide both sides of the equation by 3.
Then, the sum of the
roots will be (1/3), since we would see (-1/3) as the middle coefficient.

-(5/3) + r = 1/3
R = 6/3 = 2