Discriminant, Applications, Hidden Quadratics
The Discriminant and Factoring
If we have a quadratic expression
ax2 + bx + c
and want to determine if it factors, then we can ask an equivalent question about whether there are rational roots. In particular, there will be rational roots if the part under the square root sign in the quadratic formula is a perfect square. We call this part the discriminant which has the form
D = b2 - 4ac
3x2 + 2x - 1
can be factored.
D = 22 - 4(3)(-1) = 4 + 12 = 16
Since this is a perfect square, it can be factored. In fact, it has the factorization
3x2 + 2x - 1 = (3x -1)(x + 1)
5x2 + 8x + 2
can be factored.
D = 82 - 4(5)(2) = 64 - 40 = 24
Since this is not a perfect square, it can not be factored.
Finding the Quadratic From its Roots
The zero factor property tells us that if r is a root of a quadratic, then x - r is a factor. We can use this to reconstruct a quadratic equation from its roots.
Find a quadratic equation that has roots 2 and -3.
Since 2 and -3 are roots, we write
(x - 2) (x - (-3)) = 0
(x - 2) (x + 3) = 0
x2 + 3x - 2x - 6 = 0
x2 + x - 6 = 0
Find a quadratic expression that has roots 2 - and 2 + .
(x - (2 - ))(x - (2 + )) = 0
= ((x - 2) + )((x - 2) - ) = 0
= (x - 2)2 - 3 = 0 Difference of Squares
= x2 - 4x + 4 - 3 = 0
= x2 - 4x + 1 = 0
Find a quadratic equation that has roots
B. 4 + ,
C. 1 + i, 1 - i
Often we encounter an equation that is a quadratic in disguise. For example the equation
x4 + 3x2 - 4 = 0
is a fourth degree equation. Notice that the exponent of the leading term is twice the exponent of the second term. Because of this we can use what is called a substitution:
u = x2
x4 = u2
Then the equation in terms of u becomes:
u2 + 3u - 4 = 0
which is a quadratic. Hence, we can factor:
(u - 1)(u + 4) = 0
u = 1 or u = -4
Since u = x2 we have that
x2 = 1 or x2 = -4
The first has solution
x = 1 or x = -1
and the second has no solution.
A. x2/3 - x1/3 - 6 = 0
B. (3x2 - 4)2 + (3x2 - 4) - 2 = 0
Some pointers on solving word problems.
Avery throws a football straight up in the air. The equation
s = -16t2 + 90t
gives the distance s in feet that the football is above the ground t seconds after he throws it.
How high is the ball after 2 seconds? after 30 seconds?
How long does it take for the ball to hit the ground?
Here we are given the variable so we do not need to label it. First we calculate
s(2) = -16(2)2 + 90(2) = 116
The ball is 116 feet above the ground after 2 seconds.
Next we calculate
s(30) = -16(30)2 + 90(30) = -6300
Notice that the negative number implies that the ball has already hit the ground. Hence the ball is on the ground after 30 seconds.
For the last question, we know that when the ball hits the ground, it is 0 feet above the ground. We solve
0 = -16t2 + 90t = 2t(-8t + 45)
Since t is not zero (this is when the player throws the ball), we have
-8t + 45 = 0
-8t = -45
t = -45/8 = 5.625
It will take 5.625 seconds for the ball to hit the ground.
Two ships leave the same port at the same time. The first ship travels due North and the second travels due East. After 1 hour the ships are 25 miles apart. The ship traveling North travels 5mph faster than the ship traveling East. How fast are the ships traveling?
We form a right triangle with 25 the length of the hypotenuse.
Let x = the speed of the ship traveling East.
Then x + 5 is the speed of the ship traveling due North.
d = rt
the legs of the triangle are
(x + 5)(1)
The Pythagorean theorem tells us that
x2 + (x + 5)2 = 252
x2 + x2 + 10x + 25 = 625.
2x2 + 10x - 600 = 0
x2 + 5x - 300 = 0
(x -15)(x + 20) = 0
x = 15 or x = -20
Since x must be positive, x = 15.
The ship traveling North is traveling at 20 mph and the ship traveling East is traveling at 15 mph.