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
Solving Quadratic Equations and Inequalities
Solving Quadratic Equations and Inequalities
Standard Form: \( ax^2 + bx + c = 0 \)
Roots: Values of \( x \) that make the quadratic equal to zero.
Quadratic Inequalities: Involve solving where the expression is less than or greater than zero.
Methods:
- Factorization: Factor, solve, test intervals
- Completing the Square: Transform into vertex form, solve precisely
Example : Solve Using Factorization
Equation: \( x^2 – 7x + 12 = 0 \)
Inequality: \( x^2 – 7x + 12 < 0 \)
▶️Answer/Explanation
Factor the quadratic:
\( x^2 – 7x + 12 = (x – 3)(x – 4) \)
Roots: \( x = 3 \), \( x = 4 \)
Solve the inequality:
- Test intervals: \( x < 3 \), \( 3 < x < 4 \), \( x > 4 \)
- Only \( 3 < x < 4 \) gives a negative product
\( x \in (3, 4) \)
Example : Solve Using Completing the Square
Equation: \( x^2 + 4x + 1 = 0 \)
Inequality: \( x^2 + 4x + 1 \leq 0 \)
▶️Answer/Explanation
Complete the square:
\( x^2 + 4x + 1 = (x + 2)^2 – 3 \)
Solve:
\( (x + 2)^2 – 3 = 0 \Rightarrow (x + 2)^2 = 3 \)
\( x + 2 = \pm \sqrt{3} \Rightarrow x = -2 \pm \sqrt{3} \)
Inequality: \( (x + 2)^2 \leq 3 \Rightarrow -\sqrt{3} \leq x + 2 \leq \sqrt{3} \)
\( \Rightarrow x \in [-2 – \sqrt{3}, -2 + \sqrt{3}] \)
The Quadratic formula.
The Quadratic Formula
The quadratic formula provides the solution(s) to any quadratic equation of the form:
\( ax^2 + bx + c = 0 \)
The solutions are given by:
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Where:
- \( a, b, c \) are coefficients from the quadratic equation.
- \( b^2 – 4ac \) is the discriminant, determining the nature of the roots.
Example:
Solve Using Quadratic Formula \( 2x^2 – 3x – 5 = 0 \)
▶️Answer/Explanation
coefficients: \( a = 2 \), \( b = -3 \), \( c = -5 \).
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-5)}}{2 \cdot 2} \)
\( = \frac{3 \pm \sqrt{9 + 40}}{4} \)
\( = \frac{3 \pm \sqrt{49}}{4} \)
\( = \frac{3 \pm 7}{4} \)
Roots:
- \( x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5 \)
- \( x = \frac{3 – 7}{4} = \frac{-4}{4} = -1 \)
The Discriminant and the Nature of Roots
The Discriminant and the Nature of Roots
For a quadratic equation of the form:
\( ax^2 + bx + c = 0 \)
The discriminant is given by:
\( \Delta = b^2 – 4ac \)
The value of \( \Delta \) determines the nature of the roots:
Example: For the equation \( 3kx^2 + 2x + k = 0 \), find the possible values of \( k \) which will give:
- Two distinct real roots
- Two equal real roots
- No real roots
▶️Answer/Explanation
The quadratic equation is:
\( 3kx^2 + 2x + k = 0 \)
So,
\( a = 3k, \quad b = 2, \quad c = k \)
Discriminant:
\( \Delta = b^2 – 4ac = 2^2 – 4(3k)(k) = 4 – 12k^2 \)
Two distinct real roots
\( \Delta > 0 \Rightarrow 4 – 12k^2 > 0 \)
\( 12k^2 < 4 \Rightarrow k^2 < \frac{1}{3} \)
\( -\frac{1}{\sqrt{3}} < k < \frac{1}{\sqrt{3}} \)
Two equal real roots
\( \Delta = 0 \Rightarrow 4 – 12k^2 = 0 \)
\( 12k^2 = 4 \Rightarrow k^2 = \frac{1}{3} \)
\( k = \pm \frac{1}{\sqrt{3}} \)
No real roots
\( \Delta < 0 \Rightarrow 4 – 12k^2 < 0 \)
\( 12k^2 > 4 \Rightarrow k^2 > \frac{1}{3} \)
\( k < -\frac{1}{\sqrt{3}} \) or \( k > \frac{1}{\sqrt{3}} \)