Mathematics Magazine for Grades 1-12  

12/2003

The road uphill and the road downhill are one and the same.
 - Heraclitus (ca. 500BC)

Grade 10

Theory:

A quadratic function is any function equivalent to one of the form

f(x) = ax2 + bx + c Here are some examples of quadratic functions

       f(x) = x2 9

       f(x) = x2 + 2x +1

A quadratic equation is any equation equivalent to one of the form

ax2 + bx + c = 0

Here are some examples of quadratic equations

       x2 9 = 0

       x2 + 2x +1 = 0

Solutions from the Previous Issue:

  1. The probability that both Adrian and Brian answer a question correctly is 18 percent. The probability that one, but not both, of them answers the question correctly is 54 percent. Given that Adrian and Brian are operating independently, what is the probability that Adrian answers the question correctly, given that Adrian is more likely than Brian to answer correctly?

Solution:  6.

Let a and b denote the probabilities that Adrian and Brian, respectively, answer the question correctly. Then a∙b = 0.18 and a∙(1 - b) + (1 - a)∙b = 0.54.

The second equation simplifies to a + b - 2a∙b = 0.54.

Substituting a∙b = 0.18 gives us a + b - 2 (0.18) = 0.54, so a + b = 0.9, which implies that a + 18/a = 0.9, so a2 - 0.9a + 0.18 = 0.

Factoring the left-hand side of the equation reveals that

(a - 0.6)∙(a - 0.3) = 0, so a is either 0.6 or 0.3,

   and b is either 0.3 or 0.6, respectively.

Since a > b, we conclude that the probability that Adrian gets the answer correct is 0.6.

  1. If x2 y2 = 2xy, and if x and y are both positive, find the ratio x/y.

Solution:

Approximately 2.414.

Dividing both sides of the given equation by y2 and keeping in mind that y > 0 indicates that ; so x/y is a root of the equation w2 2w 1 = 0. Since the roots of this equation are  and since x/y must be positive, we see that ;

  1. Find all positive values of k for which the equations

3x + k = 2 and

kx + 3 = 2 have a common solution for x.

Solution: 3.

We can see that when k = 3, the equations are solved by all x.

Do other solutions exist? If such a k and x exist, then 3x + k = kx + 3.

Isolating x gives us 3x - kx = 3 - k, which is equivalent to (3 - k)x = 3 - k.

If k ≠ 3, then x must be = 1. But k(1) + 3 = 2 implies that k = -1, which is not positive, so the only positive value of k that works is k = 3.

  1. Find the sum of the coefficients in the expansion of (2a + b - c)8.

Solution: 256.

Before expanding our polynomial, we think about another way to determine the sum of its coefficients. For any polynomial f(x),  f(1) is the sum of its coefficients when expanded. Therefore, the sum of the coefficients in our polynomial is the polynomial evaluated at x = 1, that is, (2 + 1 - 1)8, or 256.

Proposed Exercises:

Use the quadratic formula to solve each equation.

  1. -2x2 - 5x + 3= 0
  2. -3x2 - 2x+ 5 = 0
  3. -x2 + 6x 5 = 0
  4. 2x2 - x - 3= 0
  5. -3x2 + 11 x - 6= 0