Mathematics Magazine for Grades 1-12
The road uphill and
the road downhill are one and the same.
A quadratic function is any function equivalent to one of the formf(x) = ax2 + bx + c Here are some examples of quadratic functions
Ø f(x) = x2 – 9
Ø 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
Ø x2 – 9 = 0
x2 + 2x +1 = 0
Solutions from the Previous Issue:
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.
Dividing both sides of the given
equation by y2 and keeping in mind that y > 0
; 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
3x + k = 2 and
kx + 3 = 2 have a common solution for x.
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.
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.
Use the quadratic formula to solve each equation.