Deriving the Quadratic Formula:
ax² + bx + c = 0
Step 1: Complete the square for ax² + bx + c = 0:
Take out a common factor of a:
a[x² + (b/a)x + c/a] = 0
Complete the square for [x² + (b/a)x + c/a]
a [ (x + b/2a)² + c/a - (b/2a)² ] = 0
Divide by a:
(x + b/2a)² + c/a - (b/2a)² = 0
Step 2: Solve the equation:
(x + b/2a)² + c/a - b²/4a² = 0
(x + b/2a)² = b²/4a² - c/a
Simplify b²/4a² - c/a:
(x + b/2a)² = (b² - 4ac) / 4ac
x + b/2a = +/- √( (b² - 4ac) / 4a² )
x + b/2a = ( +/- √(b² - 4ac) )/ 2a
x = (-b +/- √(b² - 4ac) )/ 2a
Step 1: Complete the square for ax² + bx + c = 0:
Take out a common factor of a:
a[x² + (b/a)x + c/a] = 0
Complete the square for [x² + (b/a)x + c/a]
a [ (x + b/2a)² + c/a - (b/2a)² ] = 0
Divide by a:
(x + b/2a)² + c/a - (b/2a)² = 0
Step 2: Solve the equation:
(x + b/2a)² + c/a - b²/4a² = 0
(x + b/2a)² = b²/4a² - c/a
Simplify b²/4a² - c/a:
(x + b/2a)² = (b² - 4ac) / 4ac
x + b/2a = +/- √( (b² - 4ac) / 4a² )
x + b/2a = ( +/- √(b² - 4ac) )/ 2a
x = (-b +/- √(b² - 4ac) )/ 2a
Examples:
4u² - 11u + 6 = 0
a = 4 b = -11 c = 6 u = ( - ( -11) +/- √((-11)² - (4 x 4 x 6)) )/ (2 x 4) u = (11 +/- √(121 - 96) )/ 8 u = (11 +/- √25)/ 8 u = (11 +/- 5)/ 8 u = (11 + 5)/ 8 = 16/8 = 2 u = (11 - 5)/ 8 = 6/8 = 3/4 u = 2 or u = 3/4 |
12t² - 11t - 5 = 0
a = 12 b = -11 c = -5 u = ( - ( -11) +/- √((-11)² - (4 x 12 x -5)) )/ (2 x 12) u = (11 +/- √(121 - (-240)) )/ 24 u = (11 +/- √(121 + 240) )/ 24 u = (11 +/- √361)/ 24 u = (11 +/- 19)/ 24 u = (11 + 19)/ 24 = 30/24 = 5/4 u = (11 - 19)/ 24 = -8/24 = -1/3 u = 5/4 or u = -1/3 |
The Discriminant:
- From the quadratic formula
- b² - 4ac is the discriminant
a) When b² - 4ac > 0:
- The quadratic ax² + bx + c = 0 will have two solutions.
- The curve y = ax² + bx + c will have two x intercepts.
- Therefore, the quadratic ax² + bx + c = 0 will have two distinct roots.
- The quadratic ax² + bx + c = 0 will have only one solution.
- The curve y = ax² + bx + c will have only one x intercept.
- Therefore, the quadratic ax² + bx + c = 0 will have two equal or repeated roots.
- The quadratic ax² + bx + c = 0 will have no solutions.
- The curve y = ax² + bx + c will have no x intercepts.
- Therefore, the quadratic ax² + bx + c = 0 will have no real roots.
- The quadratic ax² + bx + c = 0 can be factorised.
Examples:
Use the discriminant to calculate how many solutions the following functions have:
f(x) = 0 = 2x² - 5x - 7 f(x) = (-5)² - (4 x 2 x -7) f(x) = 25 - (-56) f(x) = 25 + 56 f(x) = 81 Therefore: f(x) has two solutions. f(x) = 0 = 9x² + 12x+ 4 f(x) = (12)² - (4 x 9 x 4) f(x) = 144 - 144 f(x) = 0 Therefore: f(x) has one solution. |
Use the discriminant to calculate how many solutions the following functions have:
f(x) = 0 = 5x² - x + 3 f(x) = (-1)² - (4 x 5 x 3) f(x) = 1 - 60 f(x) = -59 Therefore: f(x) has no solutions. f(x) = 0 = 4x² - 20x + 25 f(x) = (-20)² - (4 x 4 x 25 f(x) = 400 - 400 f(x) = 0 Therefore: f(x) has one solution. |
Examples:
Use the discriminant to calculate the value of m:
2x² - mx + 8 = 0 b² - 4ac < 0 (-m)² - (4 x 2 x 8) < 0 m² - 64 < 0 m² < 64 m < 8 |
Use the discriminant to calculate the value of a:
ax² + 16x + 32 = 0 b² - 4ac = 0 (16)² - (4 x a x 32) = 0 256 - 128a = 0 256 = 128a 2 = a |
Use the discriminant to calculate the value of p:
2x² + 6x + p = 0 b² - 4ac > 0 (6)² - (4 x 2 x p) > 0 36 - 8p > 0 36 > 8p 9/2 > p |
Examples:
Find the values of k for which x² + kx + 9 = 0 has equal roots.
b² + 4ac = 0
k² - (4 x 1 x 9) = 0
k² - 36 = 0
(k-6)(k+6) = 0
k = 6 or k = -6
b² + 4ac = 0
k² - (4 x 1 x 9) = 0
k² - 36 = 0
(k-6)(k+6) = 0
k = 6 or k = -6
Given that x² + 10x + 36 ≡ (x + a)² + b where a and b are constants:
a) Find the values of a and b
Complete the square:
x² + 10x + 36 ≡ (x+5)² + 36 - 25
x² + 10x + 36 ≡ (x+5)² + 11
b) Hence show that x² + 10x + 36 = 0
(x+5)² + 11 = 0
(x+5)² = -11
x + 5 ≠ √-11
Cannot square root a negative number.
Therefore no real roots.
Also,
y = (x+5)² = 11
Vertex = (-5, 11)
Once this graph is drawn out, it is clear to see that there are no x intercepts.
This proves that there are no real roots for this quadratic.
The equation x² + 10x + k = 0 has equal roots.
c) Find the value of k
b² - 4ac = 0
10² - (4 x 1 x k) = 0
100 - 4k = 0
100 - 4k
25 = k
d) For this value of k sketch the graph y = x² + 10x + k
y = x² + 10x + 25
y = (x+5)(x+5)
x = -5 x = -5
y intercept = 25
Graph below>>>
a) Find the values of a and b
Complete the square:
x² + 10x + 36 ≡ (x+5)² + 36 - 25
x² + 10x + 36 ≡ (x+5)² + 11
b) Hence show that x² + 10x + 36 = 0
(x+5)² + 11 = 0
(x+5)² = -11
x + 5 ≠ √-11
Cannot square root a negative number.
Therefore no real roots.
Also,
y = (x+5)² = 11
Vertex = (-5, 11)
Once this graph is drawn out, it is clear to see that there are no x intercepts.
This proves that there are no real roots for this quadratic.
The equation x² + 10x + k = 0 has equal roots.
c) Find the value of k
b² - 4ac = 0
10² - (4 x 1 x k) = 0
100 - 4k = 0
100 - 4k
25 = k
d) For this value of k sketch the graph y = x² + 10x + k
y = x² + 10x + 25
y = (x+5)(x+5)
x = -5 x = -5
y intercept = 25
Graph below>>>