IB Mathematics AA SL Use of the discriminant Study Notes
IB Mathematics AA SL Use of the discriminant Study Notes Offer a clear explanation of Use of the discriminant , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Use of the discriminant
Use of the Discriminant
The discriminant is a key component in understanding the nature of the roots of a quadratic equation. It helps determine whether the quadratic equation has real or complex roots and, if real, whether they are distinct or equal.
1. The Quadratic Equation
The general form of a quadratic equation is given by:
\( ax^2 + bx + c = 0 \)
To find the roots of this equation, we use the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
The expression inside the square root, \( \Delta = b^2 – 4ac \), is called the discriminant.
2. Nature of the Roots Based on the Discriminant
The value of the discriminant \( \Delta \) determines the nature of the roots:
- \( \Delta > 0 \): The equation has two distinct real roots.
- \( \Delta = 0 \): The equation has two equal real roots (one repeated root).
- \( \Delta < 0 \): The equation has no real roots, but two complex roots.
3. Example Problem
Consider the equation \( 3kx^2 + 2x + k = 0 \). Find the possible values of \( k \) for different types of roots:
Case 1: Two Distinct Real Roots
For two distinct real roots, we need \( \Delta > 0 \).
Calculate the discriminant: \( \Delta = (2)^2 – 4(3k)(k) = 4 – 12k^2 \)
For distinct real roots: \( 4 – 12k^2 > 0 \)
Thus, \( 1/3 > k^2 \) or \( -\sqrt{1/3} < k < \sqrt{1/3} \).
Case 2: Two Equal Real Roots
For two equal real roots, we need \( \Delta = 0 \).
Set the discriminant to zero: \( 4 – 12k^2 = 0 \)
Solving this gives \( k^2 = 1/3 \) or \( k = \pm\sqrt{1/3} \).
Case 3: No Real Roots
For no real roots, we need \( \Delta < 0 \).
Calculate \( 4 – 12k^2 < 0 \)
This implies \( k^2 > 1/3 \), so \( k < -\sqrt{1/3} \) or \( k > \sqrt{1/3} \).
4. Real-World Connections
The discriminant concept links to various real-world phenomena and other subjects:
- Physics: Used in analyzing projectile motion and energy changes in simple harmonic motion.
- Chemistry: Equilibrium equations often involve solving quadratic equations.
5. Historical Insights
Ancient mathematicians such as those from Babylon and India contributed to solving quadratic equations:
- Babylonian Method: The method of multiplication \( ab = \frac{(a+b)^2 – a^2 – b^2}{2} \).
- Sulba Sutras & Bakhshali Manuscript: These ancient texts contain algebraic methods for solving quadratic equations.
Examples of Solving Quadratic Equations Using the Discriminant
The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\( \Delta = b^2 – 4ac \)
The discriminant helps us determine the nature of the roots:
- If \( \Delta > 0 \): The equation has two distinct real roots.
- If \( \Delta = 0 \): The equation has exactly one real root (a repeated root).
- If \( \Delta < 0 \): The equation has no real roots (two complex roots).
Example 1: Discriminant Calculation with Two Distinct Real Roots
Consider the equation: \( x^2 – 5x + 6 = 0 \)
Step 1: Identify \( a = 1 \), \( b = -5 \), and \( c = 6 \).
Step 2: Calculate the discriminant:
\( \Delta = (-5)^2 – 4(1)(6) \)
\( \Delta = 25 – 24 = 1 \)
Step 3: Since \( \Delta > 0 \), the equation has two distinct real roots.
Solution: The roots can be calculated as:
\( x = \frac{-(-5) \pm \sqrt{1}}{2(1)} \)
\( x = \frac{5 \pm 1}{2} \)
\( x = 3 \) or \( x = 2 \)
Example 2: Discriminant Calculation with One Repeated Real Root
Consider the equation: \( x^2 – 4x + 4 = 0 \)
Step 1: Identify \( a = 1 \), \( b = -4 \), and \( c = 4 \).
Step 2: Calculate the discriminant:
\( \Delta = (-4)^2 – 4(1)(4) \)
\( \Delta = 16 – 16 = 0 \)
Step 3: Since \( \Delta = 0 \), the equation has one repeated real root.
Solution: The root can be calculated as:
\( x = \frac{-(-4)}{2(1)} \)
\( x = \frac{4}{2} = 2 \)
Example 3: Discriminant Calculation with No Real Roots
Consider the equation: \( 2x^2 + 3x + 5 = 0 \)
Step 1: Identify \( a = 2 \), \( b = 3 \), and \( c = 5 \).
Step 2: Calculate the discriminant:
\( \Delta = 3^2 – 4(2)(5) \)
\( \Delta = 9 – 40 = -31 \)
Step 3: Since \( \Delta < 0 \), the equation has no real roots and has two complex roots.
Solution: The roots are complex and can be written as:
\( x = \frac{-3 \pm \sqrt{-31}}{4} \)
\( x = \frac{-3 \pm i\sqrt{31}}{4} \), where \( i \) is the imaginary unit.
IB Mathematics AA Use of the Discriminant Exam Style Worked Out Questions
Question
The diagram shows the graph of the quadratic function \(f(x)=ax^2+bx+c\) , with vertex −2, 10) .
The equation f (x) = k has two solutions. One of these solutions is x = 2 .
Write down the other solution of f (x) = k . [2]
Complete the table below placing a tick (✔) to show whether the unknown parameters
a and b are positive, zero or negative. The row for c has been completed as an example. [2]
positive
zero
negative
a
b
c
✔
State the values of x for which f (x) is decreasing. [2]
▶️Answer/Explanation
Ans:
(a)
(x= (-2))-4 OR x= (-2)- (2-(-2))
X= -6
(b) 
(c) x > −2 OR x ≥ −2
Question
Let f (x)=- x2 + 4x + 5 and g (x) = -f (x) + k .
Find the values of k so that g (x) = 0 has no real roots.
▶️Answer/Explanation
Ans:
Method 1(discriminant)
correct expression for g
eg \(-(-x^{2}+4x+5)+k\) ,\( x^{2}\)-4x-5+k = 0
evidence of discriminant
eg \(b^{2}-4ac,\Delta\)
correct substitution into discriminant of g
eg \((-4)^{2}-4(1)(-5+k)\), \(16-4(k-5)\)
recognizing discriminant is negative
eg \(\Delta <0, (-4)^{2}-4(1)(-5+k)<0, 16-4 (-1)(5)<0\)
correct working (must be correct inequality)
eg -4k<-36, k-5>4,16+20-4k<0 k>9
