B10: derive and apply the quadratic formula
Originators: Kyle and Alison
Explanation: The quadratic formula is used to find values of x when y=0. These values would be the xintercepts on a graph. To derive the quadratic formula, [1]start with the equation for a quadratic function in general form and [2] let y=0. [3] Move c to the left side by subtracting c form both sides. [4] Factor a out of the right side of the equation. [5] Use the equation (b/2a)² on the terms inside the brackets and [6] add that value inside the brackets. [6] Take the same term, multiply it by the coefficient and add it to the left side or the equation. [7] Factor the brackets on the right side. [8] Make the terms on the left side of the equation one fraction and [9] move the a term to the left side by dividing both sides by a or multiply by 1/a. [10] Square root both sides and [11] move the last term on the right side to the left side to get x by itself. [12] This is the quadratic formula. To find x, substitute values for a, b, and c into this equation and solve for x.
Step 1

y = ax^{2} + bx + c

Step 2

0 = ax^{2} + bx + c

Step 3

c = ax^{2} + bx

Step 4

c = a(x^{2} + (b/a)x)

Step 5

(b/2a)^{2} = b^{2}/4a^{2}

Step 6

(b^{2}/4a) – c = a(x^{2} +(b/a)x + b^{2}/4a^{2})

Step 7

(b^{2}/4a) – c = a(x + b/2a)^{2}

Step 8

b^{2 } 4ac = a(x + b/2a)^{2}
4a

Step 9

b^{2 } 4ac = (x + b/2a)^{2}
4a^{2}

Step 10

± √(b^{2 } 4ac) = x + b/2a
2a

Step 11

b ± √(b^{2 } 4ac) = x
2a 2a

Step 12

b ± √(b^{2 } 4ac) = x
2a

Sample Question (applying the quadratic formula): While hunting, Miguel threw a rock at a delicious looking antelope. The relationship between the height of the rock in metres (y) and time in seconds (x) is represented in the quadratic equation y = 2x² + 4x + 1.
a) If the antelope 0.5 m off the ground, how long does it take for the rock to hit the antalope?
b) At what times is the rock at a height of 2 metres?
Solution Part A:
y = 2x² + 4x + 1
0.5 = 2x² + 4x + 1
0 = 2x² + 4x +0.5
x = b ± √(b^{2 } 4ac)
2a
x = 4 ± √(4^{2 }– 4(2)(0.5))
2(2)
x = 4 ± √(16^{ + }4)
4
x = 4 + √20 x = 4  √(20)
4 4
x = 0.118 x = 2.118
The rock hits the antelope after 2.118 seconds.
Solution Part B:
y = 2x² + 4x + 1
2 = 2x² + 4x + 1
0 = 2x² + 4x – 1
x = b ± √(b^{2 } 4ac)
2a
x = 4 ± √(4^{2 }– 4(2)(1))
2(2)
x = 4 ± √(16^{ } 8)
4
x = 4 ± √8
4
x = 4  √8 x = 4 + √8
4 4
x = 1.707 x = 0.293
The rock reaches a height first at .0293 seconds and again at 1.707 seconds.
Comments (1)
Anonymous said
at 8:12 pm on Mar 13, 2007
Nice work Kyle and Alison. Great sample problem.
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