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AS Level
Mathematics 9709 P1
1. Quadratics
Written by: Tharun Athreya
Formatted by: Tharun Athreya
Index
1.1 Solving quadratic equations by factorisation
1.2 Completing the square
1.3 Quadratic Equation
1.4 Solving simultaneous equations (one linear and one quadratic)
1.5 Solving more complex quadratic equations
1.6 Maximum and minimum values of a quadratic function
1.7 Solving quadratic inequalities
1.8 The number of roots of a quadratic equation
1.9 Intersection of a Line and a Quadratic Curve
1.1 Solving quadratic equations by factorisation
- A quadratic equation has the form \( ax^2 + bx + c = 0 \).
- Factorisation involves expressing the quadratic as a product of linear factors.
Methods to Factorise a Quadratic Equation
- Identify the coefficients: Find \( a, b, \) and \( c \) from the quadratic equation.
- Find the two numbers that multiply to \( ac \) and add to \( b \).
- Rewrite the middle term: Split the middle term \( bx \) into two terms using the numbers found in step 2:
\( ax^2 + bx + c = ax^2 + mx + nx + c \)
- Factor by grouping: Group the terms in pairs and factor out the common factor from each pair.
- Combine the common factors: Factor out the common factor from the grouped terms.
Example
\( x^2 + 5x + 6 = 0 \)
- Identify the coefficients: \( a = 1, b = 5, c = 6 \).
- Find two numbers that multiply to \( 6 \) and add to \( 5 \): \( 2 \) and \( 3 \).
- Rewrite the quadratic:
\( x^2 + 2x + 3x + 6 = 0 \). - Factor by grouping:
\( x(x + 2) + 3(x+2) = 0 \). - Factor out the common factors:
\( (x+2) (x+3) = 0 \).
Set each factor to zero and solve for \( x \):
\( x+2=0 \Rightarrow x=-2 \)
\( x+3=0 \Rightarrow x=-3 \)
Note: Always check your answers by expanding the factors to ensure they equal the original quadratic equation.
Key Example:
\( (x^2 – 3x + 1)^6 = 1 \)
For this question, there are three cases:
- Case 1: \( a^0 = 1 \).
- Case 2: \( 1^b = 1 \).
- Case 3: \( (-1)^b = 1 \), where \( b \) should be even.
For the key example, use:
Case 2:
- Step 1: Take the quadratic equation and equate it to 1:
\( x^2 – 3x + 1 = 1 \). - Step 2: Then solve it by bringing 1 to the other side:
\( x^2 – 3x = 0 \).
Solutions: \( x = 0, x = 3 \).
Case 3: Since 6 is an even number, we can use case 3.
- Step 1: Equate the quadratic equation to -1:
\( x^2 – 3x +1 = -1 \). - Step 2: Move -1 to the other side:
\( x^2 – 3x + 2 = 0 \). - Step 3: Solve for \( x \) using factorisation.
Solutions: \( x = 2, x = 1 \).
So, the answers are \( x = 0, x = 3, x = 1, x = 2 \).
NOTE: The number of cases needed depends on the question.
Why Are We Using Only 2 Cases?
- We use cases 2 and 3 for solving because they enable us to equate the quadratic expression to 1 and -1, respectively, which leads to real solutions.
- Case 2 allows us to set the quadratic expression to 1, providing two solutions. Case 3 applies because, with an even exponent, we can set the expression to -1, also yielding two solutions.
- Case 1, where the expression is set to zero, does not help in solving the original equation, so it is not applicable.
1.2 Completing the square
- Completing the square transforms a quadratic equation into a perfect square three-term polynomial, making it easier to solve. The goal is to rewrite the quadratic equation \( ax^2 + bx + c = 0 \) in the form of \( (x – h)^2 = k \), where \( h \) and \( k \) are constants.
Method to Complete the Square
- Write the quadratic equation as \( x^2 + bx + c \). (The coefficient of \( x^2 \) needs to be 1. If not, factor it out.)
- Determine half of the coefficient of \( x \).
- Take the square of the number obtained in the previous step.
- Add and subtract this square to the \( x^2 \) term.
- Factorise the polynomial and apply the algebraic identity:
\( x^2 + 2xy + y^2 = (x + y)^2 \) (or) \( x^2 – 2xy + y^2 = (x – y)^2 \). - General format for completing the square:
\( ax^2 + bx + c = a(x + m)^2 + n \), where \( m = \frac{b}{2a} \) and \( n = c – \frac{b^2}{4a} \).
Example
Given: \( x^2 – 4x – 8 = 0 \)
Using the formula \( ax^2 + bx + c = a(x + m)^2 + n \), we identify:
\( a = 1, \quad b = -4, \quad c = -8 \).
Calculate \( m \) and \( n \):
\( m = \frac{b}{2a} = \frac{-4}{2(1)} = -2 \)
\( n = c – \frac{b^2}{4a} = -8 – \frac{(-4)^2}{4(1)} = -12 \).
Rewriting the quadratic equation:
\( x^2 – 4x – 8 = (x – 2)^2 – 12 \).
Thus, the final form is:
\( (x – 2)^2 = 12 \).
1.3 Quadratic Equation
When solving the quadratic equation \( ax^2 + bx + c = 0 \), we use the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Example
Given equation: \( 2x^2 – 4x – 6 = 0 \)
- Identify coefficients: \( a = 2, \quad b = -4, \quad c = -6 \).
- Substituting into the formula:
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)} \)
\( x = \frac{4 \pm \sqrt{16 + 48}}{4} \)
\( x = \frac{4 \pm \sqrt{64}}{4} \)
\( x = \frac{4 \pm 8}{4} \) - Solving for \( x \):
\( x = \frac{4 + 8}{4} = 3 \) or \( x = \frac{4 – 8}{4} = -1 \).
1.4 Solving simultaneous equations (one linear and one quadratic)
- To solve simultaneous equations where one is linear and the other is quadratic, substitute the linear equation into the quadratic equation.
Solving Quadratic Simultaneous Equations
- Rearrange the linear equation so that one of the unknowns becomes the subject (if the linear equation is already in this form, you can skip to Step 2).
- Substitute the expression found in Step 1 into the quadratic equation.
- Solve the new quadratic equation from Step 2 to find the values of the unknown (there will usually be two of these).
- Substitute the values from Step 3 into the rearranged equation from Step 1 to find the values of the other unknown.
- Check your solutions by substituting the values for the two unknowns (one pair at a time!) into the original quadratic equation.
Example:
Given the equations:
\( y = x^2 + 2x + 1 \)
\( y = 3x + 5 \)
- Substituting \( y = 3x + 5 \) into \( y = x^2 + 2x + 1 \):
\( 3x + 5 = x^2 + 2x + 1 \)
- Rearrange to form a quadratic equation:
\( x^2 – x – 4 = 0 \)
- Solve the quadratic by factorization:
\( (x – 2)(x + 1) = 0 \)
- Therefore, \( x = 2 \) or \( x = -1 \).
- Substituting \( x \) back into the linear equation to find \( y \):
\( y = 3(2) + 5 = 11 \)
\( y = 3(-1) + 5 = 2 \)
- Solutions are \( (2,11) \) and \( (-1,2) \).
1.5 Solving more complex quadratic equations
- This can be said as Hidden Quadratic Equations.
Hidden Quadratic Equations
- Hidden quadratic equations are quadratics expressed in terms of a function.
Format:
\( af(x)^2 + bf(x) + c = 0 \)
(Essentially, \( x \) is replaced with a function of \( x \))
Solving Hidden Quadratic Equations
- Rearrange the equation into the format shown above.
- Replace the function of \( x \) with a new variable, solving the resulting quadratic equation.
- Substitute back the function of \( x \) into the solution to solve the original equation.
Example:
\((x – 2)^2 + 5(x – 2) + 6 = 0\)
- Identify the Hidden Quadratic Form, here \( (x – 2) \) is considered as a function of \( x \).
- Introduce a new variable:
Let \( u = x – 2 \) - Write the new equation:
\( u^2 + 5u + 6 = 0 \) - Solve the quadratic equation using factorisation:
\( (u + 2)(u + 3) = 0 \)
So \( u = -2 \) and \( u = -3 \). - Substitute back to function:
\( u = x – 2 \)
So, \( -2 = x – 2 \) and \( -3 = x – 2 \). - Solutions are \( x = 0 \) and \( x = -1 \).
1.6 Maximum and minimum values of a quadratic function
- The vertex or the turning point of a parabola represents the function’s minimum or maximum value, depending on the direction it opens. Turning point is where the gradient is zero.
Finding the vertex of a parabola
| Equation | Vertex |
|---|---|
| Standard Form: \( y = ax^2 + bx + c \) | \( \left( \frac{-b}{2a}, y \right) \) |
| Vertex Form: \( y = a (x – h)^2 + k \) | \( (h, k) \) |
Finding the Maximum and Minimum Point in a Quadratic Function
- Write your quadratic function in standard or vertex form.
- Make it the completing the square format.
- Simplify to form the new quadratic equation.
- Find the vertex using the new quadratic equation.
Find the minimum value of \( y = 2x^2 – 4x + 1 \).
- Write in vertex form:
\( y = 2(x^2 – 2x) + 1 \). - Complete the square form:
\( y = 2((x – 1)^2 – 1) + 1 \). - Simplify: \( y = 2(x – 1)^2 – 2 + 1 = 2(x – 1)^2 – 1 \).
- The vertex is at \( (1, -1) \) and since \( a > 0 \), the minimum value is \( y = -1 \).
When asked to sketch the graph of a quadratic, the key features are:
- The general shape of the graph
- The axis intercepts
- The coordinates of the vertex
1.7 Solving quadratic inequalities
- Key Point: If we multiply or divide sides of an inequality by a negative number, then the inequality sign must be reversed.
- There are four possible quadratic inequalities:
-
When the coefficient of \( x^2 \) is positive and we are looking for the less than region.
\( ax^2 + bx + c < 0 \)
Example:
-
When the coefficient of \( x^2 \) is positive and we are looking for the less than region.
-
-
When the coefficient of \( x^2 \) is positive and we are looking for the greater than region.
\( ax^2 + bx + c > 0 \)
Example:
-
When the coefficient of \( x^2 \) is positive and we are looking for the greater than region.
- When the coefficient of \( x^2 \) is negative and we are looking for the greater than region.
\(-ax^2 + bx + c > 0\)
Example:
- When the coefficient of \( x^2 \) is negative and we are looking for the greater than region.
- When the coefficient of \( x^2 \) is negative and we are looking for the less than region.
\(-ax^2 + bx + c < 0\)
Example:
- When the coefficient of \( x^2 \) is negative and we are looking for the less than region.
Example:
Solve \( x^2 – 3x – 4 > 0 \)
Solve \( x^2 – 3x – 4 > 0 \)
- Factorise the quadratic: \( (x – 4)(x + 1) > 0 \).
- Determine the roots: \( x = -1 \) and \( x = 4 \).
- Test intervals:
- For \( x < -1 \), choose \( x = -2 \): \( (-2 - 4)(-2 + 1) = 6 > 0 \).
- For \( -1 < x < 4 \), choose \( x = 0 \): \( (0 - 4)(0 + 1) = -4 < 0 \).
- For \( x > 4 \), choose \( x = 5 \): \( (5 – 4)(5 + 1) = 6 > 0 \).
- Solution: \( x < -1 \) or \( x > 4 \).
1.8 The number of roots of a quadratic equation
- For the quadratic \( ax^2 + bx + c \), the discriminant is \( b^2 – 4ac \).
- The discriminant is used to determine the number of roots that a quadratic has and the nature of those roots.
- A quadratic either has two real roots, one real root (repeated root), or no real roots.
- If a quadratic has two real roots, then:
\( b^2 – 4ac > 0 \)
- If a quadratic equation has one real root, then:
\( b^2 – 4ac = 0 \)
- If a quadratic equation has no real roots, then:
\( b^2 – 4ac < 0 \)
Example:
Determine the number of roots for \( x^2 + 4x + 5 = 0 \)
Determine the number of roots for \( x^2 + 4x + 5 = 0 \)
- Calculate the discriminant using the formula \( b^2 – 4ac \):
\( 4^2 – (4 \cdot 1 \cdot 5) = 16 – 20 = -4 \) - Since it is less than 0, there are no real roots.
1.9 Intersection of a Line and a Quadratic Curve
- The intersection points of a line and a quadratic curve are found by solving their equations simultaneously.
Example:
Find \( k \) for which \( y = x + k \) is a tangent to \( y = x^2 + 2x + 1 \).
- Set the equations equal: \( x + k = x^2 + 2x + 1 \).
- Rearrange to form a quadratic: \( x^2 + x + 1 – k = 0 \).
- For the line to be a tangent, the quadratic must have exactly one solution (discriminant \( = 0 \)):
\( 1^2 – 4(1)(1 – k) = 1 – 4 + 4k = 4k – 3 \)
- Set the discriminant to zero: \( 4k – 3 = 0 \).
- Solve for \( k \): \( k = \frac{3}{4} \).
