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Worked Examples

Q1

(i) Express \( x^2 + 8x + 15 \) in the form \( (x + a)^2 + b \).

Solution:

• Complete the square by adding and subtracting \( 16 \): \[ x^2 + 8x + 15 = (x^2 + 8x + 16 - 16) + 15 = (x + 4)^2 - 1 \]
• So the expression is: \[ (x + 4)^2 - 1 \]

Q2

(i) Express \( 2x^2 + 8x + 1 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 + 8x + 1 = 2(x^2 + 4x) + 1 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 2(x^2 + 4x + 4 - 4) + 1 = 2((x + 2)^2 - 4) + 1 \]
• Expand and simplify: \[ 2(x + 2)^2 - 8 + 1 = 2(x + 2)^2 - 7 \]
• So the expression is: \[ 2(x + 2)^2 - 7 \]

Q3

(i) Express \( x^2 - 6x + 4 \) in the form \( (x + a)^2 + b \).

Solution:

• Complete the square by adding and subtracting \( 9 \): \[ x^2 - 6x + 4 = (x^2 - 6x + 9 - 9) + 4 = (x - 3)^2 - 5 \]
• So the expression is: \[ (x - 3)^2 - 5 \]

Q4

(i) Express \( 3x^2 + 12x + 5 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 3 from the first two terms: \[ 3x^2 + 12x + 5 = 3(x^2 + 4x) + 5 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 3(x^2 + 4x + 4 - 4) + 5 = 3((x + 2)^2 - 4) + 5 \]
• Expand and simplify: \[ 3(x + 2)^2 - 12 + 5 = 3(x + 2)^2 - 7 \]
• So the expression is: \[ 3(x + 2)^2 - 7 \]

Q5

(i) Express \( x^2 + 10x + 25 \) in the form \( (x + a)^2 + b \).

Solution:

• Since \( 25 \) is already a perfect square, rewrite the expression directly: \[ x^2 + 10x + 25 = (x + 5)^2 \]
• So the expression is: \[ (x + 5)^2 \]

Q6

(i) Express \( x^2 - 4x + 4 \) in the form \( (x + a)^2 + b \).

Solution:

• Since \( 4 \) is already a perfect square, rewrite the expression directly: \[ x^2 - 4x + 4 = (x - 2)^2 \]
• So the expression is: \[ (x - 2)^2 \]

Q7

(i) Express \( 4x^2 + 16x + 9 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 4 from the first two terms: \[ 4x^2 + 16x + 9 = 4(x^2 + 4x) + 9 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 4(x^2 + 4x + 4 - 4) + 9 = 4((x + 2)^2 - 4) + 9 \]
• Expand and simplify: \[ 4(x + 2)^2 - 16 + 9 = 4(x + 2)^2 - 7 \]
• So the expression is: \[ 4(x + 2)^2 - 7 \]

Q8

(i) Express \( 5x^2 + 20x + 5 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 5 from the first two terms: \[ 5x^2 + 20x + 5 = 5(x^2 + 4x) + 5 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 5(x^2 + 4x + 4 - 4) + 5 = 5((x + 2)^2 - 4) + 5 \]
• Expand and simplify: \[ 5(x + 2)^2 - 20 + 5 = 5(x + 2)^2 - 15 \]
• So the expression is: \[ 5(x + 2)^2 - 15 \]

Q9

(i) Express \( 2x^2 + 8x + 7 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 + 8x + 7 = 2(x^2 + 4x) + 7 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 2(x^2 + 4x + 4 - 4) + 7 = 2((x + 2)^2 - 4) + 7 \]
• Expand and simplify: \[ 2(x + 2)^2 - 8 + 7 = 2(x + 2)^2 - 1 \]
• So the expression is: \[ 2(x + 2)^2 - 1 \]

Q10

(i) Express \( x^2 + 12x + 36 \) in the form \( (x + a)^2 + b \).

Solution:

• Since \( 36 \) is already a perfect square, rewrite the expression directly: \[ x^2 + 12x + 36 = (x + 6)^2 \]
• So the expression is: \[ (x + 6)^2 \]

Definitions

Linear Equation: A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. The general form of a linear equation is \( ax + b = 0 \).

Quadratic Equation: A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \).

Key Concepts and Properties

• Solutions of Linear Equations: The solution to a linear equation is the value of \( x \) that satisfies the equation. It can be found by isolating \( x \).
• Solutions of Quadratic Equations: A quadratic equation can have two solutions, which can be found by factoring, completing the square, or using the quadratic formula:

• Quadratic Formula: The solutions for the quadratic equation \( ax^2 + bx + c = 0 \) are given by:

  \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• Discriminant: The discriminant \( \Delta = b^2 - 4ac \) determines the nature of the roots:
  • If \( \Delta > 0 \), there are two distinct real roots.
  • If \( \Delta = 0 \), there is exactly one real root.
  • If \( \Delta < 0 \), there are no real roots (the solutions are complex).

Worked Examples

Example 1: Solving a Linear Equation

Solve the equation \( 2x + 3 = 7 \).

Solution: Subtract 3 from both sides:

• \(2x = 4\)

• Then, divide by 2:

  \(x = 2\)

Example 2:

Example of solving the equation \(5(x - 3) = 2(x + 6)\).

Worked Examples

Q1 (Solving Quadratic Equation using the Quadratic Formula)

Solve the quadratic equation \( 3x^2 - 4x - 7 = 0 \) using the quadratic formula.

Solution:

• Start with the given quadratic equation: \[ 3x^2 - 4x - 7 = 0 \]
• The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• For the equation \( 3x^2 - 4x - 7 = 0 \), the coefficients are: \[ a = 3, \quad b = -4, \quad c = -7 \]
• Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-7)}}{2(3)} \]
• Simplify the expression: \[ x = \frac{4 \pm \sqrt{16 + 84}}{6} \]
• Simplify further: \[ x = \frac{4 \pm \sqrt{100}}{6} \]
• The square root of 100 is 10: \[ x = \frac{4 \pm 10}{6} \]
• Now, calculate the two possible values of \( x \):
  • For the positive root: \[ x = \frac{4 + 10}{6} = \frac{14}{6} = \frac{7}{3} \]
  • For the negative root: \[ x = \frac{4 - 10}{6} = \frac{-6}{6} = -1 \]
• Thus, the solutions to the quadratic equation are: \[ x = \frac{7}{3} \quad \text{or} \quad x = -1 \]

Q2 (Application of the Quadratic Formula to a Real-Life Problem)

A ball is thrown upwards from a height of 2 meters with an initial velocity of 10 meters per second. The height \( h \) of the ball after \( t \) seconds is given by the equation \( h = -5t^2 + 10t + 2 \). Find the time when the ball hits the ground.

Solution:

• The ball hits the ground when the height \( h \) is 0. Set \( h = 0 \) in the equation: \[ -5t^2 + 10t + 2 = 0 \]
• This is a quadratic equation in \( t \), so we can solve it using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• For the equation \( -5t^2 + 10t + 2 = 0 \), the coefficients are: \[ a = -5, \quad b = 10, \quad c = 2 \]
• Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ t = \frac{-10 \pm \sqrt{(10)^2 - 4(-5)(2)}}{2(-5)} \]
• Simplify the expression: \[ t = \frac{-10 \pm \sqrt{100 + 40}}{-10} \]
• Simplify further: \[ t = \frac{-10 \pm \sqrt{140}}{-10} \]
• Since \( \sqrt{140} \approx 11.83 \), we have: \[ t = \frac{-10 \pm 11.83}{-10} \]
• Now, calculate the two possible values of \( t \):
  • For the positive root: \[ t = \frac{-10 + 11.83}{-10} \approx \frac{1.83}{-10} \approx -0.183 \] (This is not physically possible, as time cannot be negative, so we discard this solution.)
  • For the negative root: \[ t = \frac{-10 - 11.83}{-10} \approx \frac{-21.83}{-10} \approx 2.183 \]
• Thus, the ball hits the ground after approximately \( 2.18 \) seconds.

Solving Quadratic Equation using Factorization

Example 1

Solve the quadratic equation \( 2x^2 + 7x + 3 = 0 \) using the factorization method.

Solution:

• Start with the given quadratic equation: \[ 2x^2 + 7x + 3 = 0 \]
• We need to factorize the quadratic expression on the left-hand side.
• First, find two numbers that multiply to give the product of the coefficient of \( x^2 \) (which is 2) and the constant term (which is 3), i.e.: \[ 2 \times 3 = 6 \]
• Next, we need two numbers that multiply to give 6 and add to give the coefficient of \( x \), which is 7. These numbers are 6 and 1 because: \[ 6 \times 1 = 6 \quad \text{and} \quad 6 + 1 = 7 \]
• Now, rewrite the middle term (7x) as a sum of 6x and 1x: \[ 2x^2 + 6x + x + 3 = 0 \]
• Group the terms in pairs: \[ (2x^2 + 6x) + (x + 3) = 0 \]
• Factor out the greatest common factor (GCF) from each group: \[ 2x(x + 3) + 1(x + 3) = 0 \]
• Factor out the common binomial factor: \[ (2x + 1)(x + 3) = 0 \]
• Set each factor equal to zero and solve for \( x \):
  • For \( 2x + 1 = 0 \): \[ 2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2} \]
  • For \( x + 3 = 0 \): \[ x = -3 \]
• Thus, the solutions to the quadratic equation are: \[ x = -\frac{1}{2} \quad \text{and} \quad x = -3 \]

Example 2 (Application of Factorization to Solve a Real-Life Problem)

A rectangle has an area of 24 square meters. Its length is 2 meters longer than its width. Find the dimensions of the rectangle.

Solution:

• Let the width of the rectangle be \( x \) meters. Then, the length of the rectangle is \( x + 2 \) meters.
• The area of the rectangle is given by: \[ \text{Area} = \text{Length} \times \text{Width} \] \li>Substitute the values for area and dimensions: \[ 24 = (x + 2) \times x \]
• Expand the equation: \[ 24 = x^2 + 2x \]
• Rearrange the equation to form a quadratic equation: \[ x^2 + 2x - 24 = 0 \]
• Factorize the quadratic equation. Find two numbers that multiply to give -24 and add to give 2. These numbers are 6 and -4 because: \[ 6 \times (-4) = -24 \quad \text{and} \quad 6 + (-4) = 2 \]
• Rewrite the middle term as a sum of these two numbers: \[ x^2 + 6x - 4x - 24 = 0 \]
• Group the terms in pairs and factorize: \[ (x + 6)(x - 4) = 0 \]
• Set each factor equal to zero:
  • For \( x + 6 = 0 \): \[ x = -6 \]
  • For \( x - 4 = 0 \): \[ x = 4 \]
• Since the width of the rectangle cannot be negative, the width is \( x = 4 \) meters.
• The length of the rectangle is: \[ x + 2 = 4 + 2 = 6 \quad \text{meters} \]
• Thus, the dimensions of the rectangle are 4 meters by 6 meters.

Worked Examples

Q1 (Completing the Square)

Express the quadratic expression \( x^2 + 6x + 5 \) in the form \( (x + p)^2 + q \).

Solution:

• Start with the given quadratic expression: \[ x^2 + 6x + 5 \]
• The goal is to rewrite it as a perfect square plus or minus a constant. We focus on the first two terms, \( x^2 + 6x \).
• To complete the square, take half of the coefficient of \( x \) (which is 6), divide it by 2 to get 3, and then square it to get 9.
• Add and subtract 9 to the expression: \[ x^2 + 6x + 9 - 9 + 5 \]
• The first three terms now form a perfect square: \[ (x + 3)^2 - 9 + 5 \]
• Simplify the constants: \[ (x + 3)^2 - 4 \]
• Thus, the quadratic expression \( x^2 + 6x + 5 \) can be written in the form: \[ (x + 3)^2 - 4 \]

Q2 (Completing the Square to Solve a Quadratic Equation)

Solve the quadratic equation \( x^2 + 6x + 5 = 0 \) by completing the square.

Solution:

• We have already expressed the quadratic as: \[ (x + 3)^2 - 4 \]
• Set the equation equal to zero: \[ (x + 3)^2 - 4 = 0 \]
• Add 4 to both sides: \[ (x + 3)^2 = 4 \]
• Take the square root of both sides: \[ x + 3 = \pm 2 \]
• Solve for \( x \): \[ x = -3 \pm 2 \]
• So, the two solutions are: \[ x = -1 \quad \text{or} \quad x = -5 \]

Q3 (Application of Completing the Square)

Find the minimum value of the quadratic function \( f(x) = x^2 + 6x + 5 \) and state the value of \( x \) at which it occurs.

Solution:

• From the previous work, we know that the expression can be written as: \[ f(x) = (x + 3)^2 - 4 \]
• The minimum value of \( (x + 3)^2 \) is zero, which occurs when \( x = -3 \).
• Therefore, the minimum value of \( f(x) \) is: \[ f(x) = 0 - 4 = -4 \]
• Thus, the minimum value of the function is \( -4 \), and it occurs at \( x = -3 \).

Definitions

Linear Equation: A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. The general form of a linear equation is \( ax + b = 0 \).

Quadratic Equation: A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \).

Vocabulary List

• General Form: The standard format of an equation, such as \( ax + b = 0 \) for linear equations and \( ax^2 + bx + c = 0 \) for quadratic equations.
• Solution: The value(s) of the variable that satisfy the equation.
• Constant: A fixed value that does not change.
• Variable: A symbol, usually a letter, that represents an unknown value.
• Factoring: The process of breaking down a quadratic equation into simpler expressions that multiply together to give the original equation.
• Quadratic Formula: A formula that provides the solutions to a quadratic equation.
• Discriminant: The expression under the square root in the quadratic formula, which determines the nature of the roots.
• Roots: The solutions to a quadratic equation; also known as zeros of the equation.
• Real Roots: The solutions to a quadratic equation that are real numbers.
• Complex Roots: The solutions to a quadratic equation that are complex numbers.
• Isolating: The process of rearranging an equation to get the variable of interest by itself on one side of the equation.
• Degree of an Equation: The highest power of the variable in the equation.
• Binomial: An algebraic expression containing two terms.
• Factorization: The process of breaking down an expression into a product of simpler expressions.
• Real-Life Application: The use of mathematical concepts, such as equations, to solve problems in everyday situations.
• Parabola: The graphical representation of a quadratic equation, showing the curve created by the equation.
• Intercept: The point where the graph of an equation crosses an axis.
• Standard Form: A specific arrangement of an equation or expression, particularly used in quadratic equations as \( ax^2 + bx + c = 0 \).
• Negative Discriminant: Indicates that a quadratic equation has no real roots but has two complex roots.
• Positive Discriminant: Indicates that a quadratic equation has two distinct real roots.
• Zero Discriminant: Indicates that a quadratic equation has exactly one real root, which is a repeated root.
• Vertex: The highest or lowest point on the graph of a parabola, representing the maximum or minimum value of the quadratic function.
• Axis of Symmetry: The vertical line that divides the parabola into two mirror-image halves, typically given by \( x = -\frac{b}{2a} \) for a quadratic equation.

Algebra Week 1 Teaching Plan

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Linear Equations Self Assessment|Reflection Quiz

Quadratic Equations QuizInstruction : Enter both roots separated by a comma

Worked Examples

Q1 (Jun 2005, Q2)

(i) Express \( 3x^2 + 12x + 7 \) in the form \( 3(x + a)^2 + b \).

Solution:

• Factor out the 3 from the first two terms: \[ 3x^2 + 12x + 7 = 3(x^2 + 4x) + 7 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 3(x^2 + 4x + 4 - 4) + 7 = 3((x + 2)^2 - 4) + 7 \]
• Expand and simplify: \[ 3(x + 2)^2 - 12 + 7 = 3(x + 2)^2 - 5 \]
• So the expression is: \[ 3(x + 2)^2 - 5 \]

(ii) Hence write down the equation of the line of symmetry of the curve \( y = 3x^2 + 12x + 7 \).

Solution:

• The equation of the line of symmetry is \( x = -a \) where \( a = 2 \), so: \[ x = -2 \]

Q2 (Jun 2006, Q3)

(i) Express \( 2x^2 + 12x + 13 \) in the form \( a(x + b)^2 + c \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 + 12x + 13 = 2(x^2 + 6x) + 13 \]
• Complete the square by adding and subtracting \( 9 \) inside the bracket: \[ 2(x^2 + 6x + 9 - 9) + 13 = 2((x + 3)^2 - 9) + 13 \]
• Expand and simplify: \[ 2(x + 3)^2 - 18 + 13 = 2(x + 3)^2 - 5 \]
• So the expression is: \[ 2(x + 3)^2 - 5 \]

(ii) Solve \( 2x^2 + 12x + 13 = 0 \), giving your answers in simplified surd form.

Solution:

• Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• Substitute \( a = 2 \), \( b = 12 \), and \( c = 13 \): \[ x = \frac{-12 \pm \sqrt{12^2 - 4(2)(13)}}{4} = \frac{-12 \pm \sqrt{144 - 104}}{4} = \frac{-12 \pm \sqrt{40}}{4} \]
• Simplify: \[ x = \frac{-12 \pm 2\sqrt{10}}{4} = \frac{-6 \pm \sqrt{10}}{2} \]

Q3 (Jan 2007, Q6)

(i) Express \( 2x^2 - 24x + 80 \) in the form \( a(x - b)^2 + c \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 - 24x + 80 = 2(x^2 - 12x) + 80 \]
• Complete the square by adding and subtracting \( 36 \) inside the bracket: \[ 2(x^2 - 12x + 36 - 36) + 80 = 2((x - 6)^2 - 36) + 80 \]
• Expand and simplify: \[ 2(x - 6)^2 - 72 + 80 = 2(x - 6)^2 + 8 \]
• So the expression is: \[ 2(x - 6)^2 + 8 \]

(ii) State the equation of the line of symmetry of the curve \( y = 2x^2 - 24x + 80 \).

Solution:

• The equation of the line of symmetry is \( x = b \), so: \[ x = 6 \]

(iii) State the equation of the tangent to the curve \( y = 2x^2 - 24x + 80 \) at its minimum point.

Solution:

• At the minimum point, the tangent is horizontal, so its equation is: \[ y = \text{constant} = 8 \]

Q4 (Jun 2008, Q10) [Modified]

(i) Express \( 2x^2 - 6x + 11 \) in the form \( p(x + q)^2 + r \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 - 6x + 11 = 2(x^2 - 3x) + 11 \]
• Complete the square by adding and subtracting \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \) inside the bracket: \[ 2\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) + 11 = 2\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 11 \]
• Expand and simplify: \[ 2\left(x - \frac{3}{2}\right)^2 - \frac{9}{2} + 11 = 2\left(x - \frac{3}{2}\right)^2 + \frac{13}{2} \]
• So the expression is: \[ 2\left(x - \frac{3}{2}\right)^2 + \frac{13}{2} \]

(ii) State the coordinates of the vertex of the curve \( y = 2x^2 - 6x + 11 \).

Solution:

• The vertex is at \( \left(\frac{3}{2}, \frac{13}{2}\right) \).

Q5 (Jan 2009, Q6) [Modified]

(i) Express \( 5x^2 + 20x - 8 \) in the form \( p(x + q)^2 + r \).

Solution:

• Factor out the 5 from the first two terms: \[ 5x^2 + 20x - 8 = 5(x^2 + 4x) - 8 \]
• Complete the square by adding and subtracting \( 4 \) inside the bracket: \[ 5(x^2 + 4x + 4 - 4) - 8 = 5((x + 2)^2 - 4) - 8 \]
• Expand and simplify: \[ 5(x + 2)^2 - 20 - 8 = 5(x + 2)^2 - 28 \]
• So the expression is: \[ 5(x + 2)^2 - 28 \]

(ii) State the equation of the line of symmetry of the curve \( y = 5x^2 + 20x - 8 \).

Solution:

• The equation of the line of symmetry is \( x = -q \), where \( q = 2 \), so: \[ x = -2 \]

Q6 (Jun 2012, Q4)

(i) Express \( 2x^2 - 20x + 49 \) in the form \( p(x - q)^2 + r \).

Solution:

• Factor out the 2 from the first two terms: \[ 2x^2 - 20x + 49 = 2(x^2 - 10x) + 49 \]
• Complete the square by adding and subtracting \( 25 \) inside the bracket: \[ 2(x^2 - 10x + 25 - 25) + 49 = 2((x - 5)^2 - 25) + 49 \]
• Expand and simplify: \[ 2(x - 5)^2 - 50 + 49 = 2(x - 5)^2 - 1 \]
• So the expression is: \[ 2(x - 5)^2 - 1 \]

(ii) State the coordinates of the vertex of the curve \( y = 2x^2 - 20x + 49 \).

Solution:

• The vertex is at \( \left(5, -1\right) \).

Worked Examples

Q1 (Specimen Q3)

The quadratic equation \( x^2 + kx + k = 0 \) has no real roots for \( x \).

(i) Write down the discriminant of \( x^2 + kx + k \) in terms of \( k \).

Solution:

• The discriminant \( \Delta \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \]
• For the given equation \( x^2 + kx + k = 0 \), \( a = 1 \), \( b = k \), and \( c = k \): \[ \Delta = k^2 - 4(1)(k) = k^2 - 4k \]

(ii) Hence find the set of values that \( k \) can take.

Solution:

• Since the equation has no real roots, the discriminant must be less than zero: \[ k^2 - 4k < 0 \]
• Factor the quadratic inequality: \[ k(k - 4) < 0 \]
• The solution for this inequality is: \[ 0 < k < 4 \]

Q2 (Jun 2007, Q4)

(i) Find the discriminant of \( kx^2 - 4x + k \) in terms of \( k \).

Solution:

• The discriminant \( \Delta \) is given by: \[ \Delta = b^2 - 4ac \]
• For the given equation \( kx^2 - 4x + k = 0 \), \( a = k \), \( b = -4 \), and \( c = k \): \[ \Delta = (-4)^2 - 4(k)(k) = 16 - 4k^2 \]

(ii) The quadratic equation \( kx^2 - 4x + k = 0 \) has equal roots. Find the possible values of \( k \).

Solution:

• For the equation to have equal roots, the discriminant must be zero: \[ 16 - 4k^2 = 0 \]
• Solve for \( k \): \[ 4k^2 = 16 \quad \Rightarrow \quad k^2 = 4 \quad \Rightarrow \quad k = \pm 2 \]

Q3 (Jan 2010, Q10)

The quadratic equation \( kx^2 - 30x + 25k = 0 \) has equal roots. Find the possible values of \( k \).

Solution:

• The discriminant \( \Delta \) for the given equation is: \[ \Delta = (-30)^2 - 4(k)(25k) = 900 - 100k^2 \]
• For the equation to have equal roots, the discriminant must be zero: \[ 900 - 100k^2 = 0 \]
• Solve for \( k \): \[ 100k^2 = 900 \quad \Rightarrow \quad k^2 = 9 \quad \Rightarrow \quad k = \pm 3 \]

Q4 (Jan 2013, Q8)

The quadratic equation \( kx^2 + (3k - 1)x - 4 = 0 \) has no real roots. Find the set of possible values of \( k \).

Solution:

• The discriminant \( \Delta \) for the given equation is: \[ \Delta = (3k - 1)^2 - 4(k)(-4) = (9k^2 - 6k + 1) + 16k = 9k^2 + 10k + 1 \]
• Since the equation has no real roots, the discriminant must be less than zero: \[ 9k^2 + 10k + 1 < 0 \]
• Solving the quadratic inequality: \[ k = \frac{-10 \pm \sqrt{(10)^2 - 4(9)(1)}}{2(9)} = \frac{-10 \pm 8}{18} \]
• This gives two roots: \[ k = \frac{-1}{9} \quad \text{and} \quad k = -1 \]
• The solution is: \[ -1 < k < -\frac{1}{9} \]

Q5 (Jun 2015, Q8)

(i) Sketch the curve \( y = 2x^2 - x - 3 \), giving the coordinates of all points of intersection with the axes.

Solution:

• To find the points of intersection with the x-axis, solve \( y = 0 \): \[ 2x^2 - x - 3 = 0 \]
• Use the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-3)}}{2(2)} = \frac{1 \pm \sqrt{25}}{4} \]
• This gives two roots: \[ x = 2 \quad \text{and} \quad x = -\frac{3}{2} \]
• So the points of intersection with the x-axis are \( (2, 0) \) and \( \left(-\frac{3}{2}, 0\right) \).
• To find the point of intersection with the y-axis, set \( x = 0 \): \[ y = -3 \]
• So the point of intersection with the y-axis is \( (0, -3) \).

(ii) Hence, or otherwise, solve the inequality \( 2x^2 - x - 3 > 0 \).

Solution:

• The curve \( y = 2x^2 - x - 3 \) is above the x-axis when \( x < -\frac{3}{2} \) or \( x > 2 \), so the solution is: \[ x < -\frac{3}{2} \quad \text{or} \quad x > 2 \]

(iii) Given that the equation \( 2x^2 - x - 3 = k \) has no real roots, find the set of possible values of the constant \( k \).

Solution:

• The discriminant \( \Delta \) for the equation \( 2x^2 - x - (3+k) = 0 \) must be less than zero for no real roots: \[ \Delta = (-1)^2 - 4(2)(-3-k) = 1 + 8(3+k) = 1 + 24 + 8k = 25 + 8k \]
• For no real roots, we require: \[ 25 + 8k < 0 \]
• Solving for \( k \): \[ k < -\frac{25}{8} \]

Worked Examples

Q1 (Jun 2006, Q6)

Solve the inequality \( x^2 + 2x < 3 \).

Solution:

• Rearrange the inequality to standard quadratic form: \[ x^2 + 2x - 3 < 0 \]
• Factorize the quadratic expression: \[ (x + 3)(x - 1) < 0 \]
• Determine the critical points, where \( x + 3 = 0 \) or \( x - 1 = 0 \): \[ x = -3 \quad \text{and} \quad x = 1 \]
• Analyze the sign of \( (x + 3)(x - 1) \) in the intervals defined by the critical points \( x = -3 \) and \( x = 1 \):
  • For \( x < -3 \), both factors are negative, so the product is positive.
  • For \( -3 < x < 1 \), \( x + 3 \) is positive and \( x - 1 \) is negative, so the product is negative.
  • For \( x > 1 \), both factors are positive, so the product is positive.
• The inequality holds in the interval where the product is negative: \[ -3 < x < 1 \]

Q2 (Jun 2009, Q4)

Solve the inequality \( x(x - 6) > 0 \).

Solution:

• Find the critical points by solving \( x(x - 6) = 0 \): \[ x = 0 \quad \text{or} \quad x = 6 \]
• Analyze the sign of \( x(x - 6) \) in the intervals defined by \( x = 0 \) and \( x = 6 \):
  • For \( x < 0 \), both \( x \) and \( x - 6 \) are negative, so the product is positive.
  • For \( 0 < x < 6 \), \( x \) is positive and \( x - 6 \) is negative, so the product is negative.
  • For \( x > 6 \), both \( x \) and \( x - 6 \) are positive, so the product is positive.
• The inequality holds where the product is positive: \[ x < 0 \quad \text{or} \quad x > 6 \]

Q3 (Jan 2010, Q2)

Solve the inequality \( \frac{5x - 3}{2} \leq x + 5 \).

Solution:

• First, eliminate the fraction by multiplying both sides by 2: \[ 5x - 3 \leq 2(x + 5) \]
• Expand the right-hand side: \[ 5x - 3 \leq 2x + 10 \]
• Move all terms involving \( x \) to one side and constants to the other side: \[ 5x - 2x \leq 10 + 3 \] \[ 3x \leq 13 \]
• Divide both sides by 3: \[ x \leq \frac{13}{3} \]
• Thus, the solution is: \[ x \leq \frac{13}{3} \]

Q4 (Jan 2013, Q4)

Solve the inequality \( 5x^2 - 28x - 12 \leq 0 \).

Solution:

• First, solve the equation \( 5x^2 - 28x - 12 = 0 \) using the quadratic formula: \[ x = \frac{-(-28) \pm \sqrt{(-28)^2 - 4(5)(-12)}}{2(5)} \] \[ x = \frac{28 \pm \sqrt{784 + 240}}{10} \] \[ x = \frac{28 \pm \sqrt{1024}}{10} \] \[ x = \frac{28 \pm 32}{10} \]
• The two solutions are: \[ x = \frac{28 + 32}{10} = 6 \quad \text{or} \quad x = \frac{28 - 32}{10} = -0.4 \]
• So the critical points are \( x = -0.4 \) and \( x = 6 \).
• Analyze the sign of \( 5x^2 - 28x - 12 \) in the intervals defined by these critical points:
  • For \( x < -0.4 \), the expression is positive.
  • For \( -0.4 \leq x \leq 6 \), the expression is non-positive.
  • For \( x > 6 \), the expression is positive.
• Thus, the solution is: \[ -0.4 \leq x \leq 6 \]

Q5 (Jun 2014, Q6)

Solve the inequality \( 3x^2 + 10x + 3 > 0 \).

Solution:

• First, solve the equation \( 3x^2 + 10x + 3 = 0 \) using the quadratic formula: \[ x = \frac{-10 \pm \sqrt{(10)^2 - 4(3)(3)}}{2(3)} \] \[ x = \frac{-10 \pm \sqrt{100 - 36}}{6} \] \[ x = \frac{-10 \pm \sqrt{64}}{6} \] \[ x = \frac{-10 \pm 8}{6} \]
• The two solutions are: \[ x = \frac{-10 + 8}{6} = \frac{-2}{6} = -\frac{1}{3} \quad \text{or} \quad x = \frac{-10 - 8}{6} = \frac{-18}{6} = -3 \]
• So the critical points are \( x = -3 \) and \( x = -\frac{1}{3} \).
• Analyze the sign of \( 3x^2 + 10x + 3 \) in the intervals defined by these critical points:
  • For \( x < -3 \), the expression is positive.
  • For \( -3 < x < -\frac{1}{3} \), the expression is negative.
  • For \( x > -\frac{1}{3} \), the expression is positive.
• Thus, the solution is: \[ x < -3 \quad \text{or} \quad x > -\frac{1}{3} \]

Q7 (OCR 4721, Jun 2005, Q8)

The length of a rectangular children’s playground is 10 m more than its width. The width of the playground is \( x \) meters.

(i) The perimeter of the playground is greater than 64 m. Write down a linear inequality in \( x \).

Solution:

• The length is \( x + 10 \) meters.
• The perimeter of a rectangle is given by: \[ \text{Perimeter} = 2(\text{Length} + \text{Width}) = 2(x + x + 10) = 4x + 20 \]
• Set up the inequality for the perimeter: \[ 4x + 20 > 64 \]
• Simplify the inequality: \[ 4x > 44 \quad \Rightarrow \quad x > 11 \]

(ii) The area of the playground is less than 299 m². Show that \( (x - 13)(x + 23) < 0 \).

Solution:

• The area of a rectangle is given by: \[ \text{Area} = \text{Length} \times \text{Width} = (x + 10)x = x^2 + 10x \]
• Set up the inequality for the area: \[ x^2 + 10x < 299 \]
• Rearrange the inequality to standard quadratic form: \[ x^2 + 10x - 299 < 0 \]
• Factorize the quadratic expression: \[ (x - 13)(x + 23) < 0 \]

(iii) By solving the inequalities in parts (i) and (ii), determine the set of possible values of \( x \).

Solution:

• From part (i), \( x > 11 \).
• From part (ii), \( -23 < x < 13 \).
• Combining these two results, the set of possible values of \( x \) is: \[ 11 < x < 13 \]

Worked Examples

Q1 (Jun 2010, Q7)

Solve the simultaneous equations:

• \( x + 2y - 6 = 0 \)
• \( 2x^2 + y^2 = 57 \)

Solution:

• From the first equation, solve for \( x \) in terms of \( y \): \[ x = 6 - 2y \]
• Substitute \( x = 6 - 2y \) into the second equation: \[ 2(6 - 2y)^2 + y^2 = 57 \]
• Expand and simplify: \[ 2(36 - 24y + 4y^2) + y^2 = 57 \] \[ 72 - 48y + 8y^2 + y^2 = 57 \] \[ 9y^2 - 48y + 72 = 57 \] \] \[ 9y^2 - 48y + 15 = 0 \]
• Solve the quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 9 \), \( b = -48 \), and \( c = 15 \).
• Substitute the values: \[ y = \frac{48 \pm \sqrt{(-48)^2 - 4(9)(15)}}{2(9)} \] \[ y = \frac{48 \pm \sqrt{2304 - 540}}{18} \] \[ y = \frac{48 \pm \sqrt{1764}}{18} \] \[ y = \frac{48 \pm 42}{18} \]
• Thus, the two solutions for \( y \) are: \[ y = \frac{48 + 42}{18} = \frac{90}{18} = 5 \quad \text{or} \quad y = \frac{48 - 42}{18} = \frac{6}{18} = \frac{1}{3} \]
• For \( y = 5 \), substitute into \( x = 6 - 2y \): \[ x = 6 - 2(5) = -4 \]
• For \( y = \frac{1}{3} \), substitute into \( x = 6 - 2y \): \[ x = 6 - 2\left(\frac{1}{3}\right) = 6 - \frac{2}{3} = \frac{16}{3} \]
• Thus, the solutions are: \[ (x, y) = (-4, 5) \quad \text{or} \quad \left(\frac{16}{3}, \frac{1}{3}\right) \]

Q2 (Jun 2011, Q4)

Solve the simultaneous equations:

• \( y = 2(x - 2)^2 \)
• \( 3x + y = 26 \)

Solution:

• Substitute \( y = 2(x - 2)^2 \) into the second equation: \[ 3x + 2(x - 2)^2 = 26 \]
• Expand and simplify: \[ 3x + 2(x^2 - 4x + 4) = 26 \] \[ 3x + 2x^2 - 8x + 8 = 26 \] \[ 2x^2 - 5x + 8 = 26 \] \[ 2x^2 - 5x - 18 = 0 \]
• Solve the quadratic equation using the quadratic formula: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-18)}}{2(2)} \] \[ x = \frac{5 \pm \sqrt{25 + 144}}{4} \] \[ x = \frac{5 \pm \sqrt{169}}{4} \] \[ x = \frac{5 \pm 13}{4} \]
• Thus, the two solutions for \( x \) are: \[ x = \frac{5 + 13}{4} = \frac{18}{4} = 4.5 \quad \text{or} \quad x = \frac{5 - 13}{4} = \frac{-8}{4} = -2 \]
• For \( x = 4.5 \), substitute into \( y = 2(x - 2)^2 \): \[ y = 2(4.5 - 2)^2 = 2(2.5)^2 = 2(6.25) = 12.5 \]
• For \( x = -2 \), substitute into \( y = 2(x - 2)^2 \): \[ y = 2(-2 - 2)^2 = 2(-4)^2 = 2(16) = 32 \]
• Thus, the solutions are: \[ (x, y) = (4.5, 12.5) \quad \text{or} \quad (-2, 32) \]

Q3 (Jan 2013, Q4)

(i) Solve the simultaneous equations:

• \( y = 2x^2 - 3x - 5 \)
• \( 10x + 2y + 11 = 0 \)

Solution:

• Substitute \( y = 2x^2 - 3x - 5 \) into the second equation: \[ 10x + 2(2x^2 - 3x - 5) + 11 = 0 \]
• Expand and simplify: \[ 10x + 4x^2 - 6x - 10 + 11 = 0 \] \[ 4x^2 + 4x + 1 = 0 \]
• Factorize the quadratic equation: \[ (2x + 1)^2 = 0 \] \li>Solve for \( x \): \[ 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]
• Substitute \( x = -\frac{1}{2} \) into \( y = 2x^2 - 3x - 5 \): \[ y = 2\left(\frac{-1}{2}\right)^2 - 3\left(\frac{-1}{2}\right) - 5 = 2\left(\frac{1}{4}\right) + \frac{3}{2} - 5 = \frac{1}{2} + \frac{3}{2} - 5 = -3 \]
• Thus, the solution is: \[ (x, y) = \left( -\frac{1}{2}, -3 \right) \]

(ii) What can you deduce from the answer to part (i) about the curve \( y = 2x^2 - 3x - 5 \) and the line \( 10x + 2y + 11 = 0 \)?

Solution:

• The curve and the line intersect at exactly one point, \( \left( -\frac{1}{2}, -3 \right) \), which means the line is a tangent to the curve at this point.

Q4 (Jun 2015, Q6)

Solve the simultaneous equations:

• \( 2x + y - 5 = 0 \)
• \( x^2 - y^2 = 3 \)

Solution:

• From the first equation, solve for \( y \) in terms of \( x \): \[ y = 5 - 2x \]
• Substitute \( y = 5 - 2x \) into the second equation: \[ x^2 - (5 - 2x)^2 = 3 \]
• Expand and simplify: \[ x^2 - (25 - 20x + 4x^2) = 3 \] \[ x^2 - 25 + 20x - 4x^2 = 3 \] \[ -3x^2 + 20x - 28 = 0 \] \li>Multiply through by \(-1\): \[ 3x^2 - 20x + 28 = 0 \]
• Solve the quadratic equation using the quadratic formula: \[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(3)(28)}}{2(3)} \]
• Solve the quadratic equation using the quadratic formula: \[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(3)(28)}}{2(3)} \] \[ x = \frac{20 \pm \sqrt{400 - 336}}{6} \] \[ x = \frac{20 \pm \sqrt{64}}{6} \] \[ x = \frac{20 \pm 8}{6} \]
• Thus, the two solutions for \( x \) are:
  • For the positive root: \[ x = \frac{20 + 8}{6} = \frac{28}{6} = \frac{14}{3} \]
  • For the negative root: \[ x = \frac{20 - 8}{6} = \frac{12}{6} = 2 \]
• For \( x = \frac{14}{3} \), substitute into \( y = 5 - 2x \): \[ y = 5 - 2\left(\frac{14}{3}\right) = 5 - \frac{28}{3} = \frac{15}{3} - \frac{28}{3} = -\frac{13}{3} \]
• For \( x = 2 \), substitute into \( y = 5 - 2x \): \[ y = 5 - 2(2) = 5 - 4 = 1 \]
• Thus, the solutions are: \[ (x, y) = \left(\frac{14}{3}, -\frac{13}{3}\right) \quad \text{or} \quad (2, 1) \]

Q5 (Jun 2016, Q3)

Solve the simultaneous equations:

• \( x^2 + y^2 = 34 \)
• \( 3x - y + 4 = 0 \)

Solution:

• From the second equation, solve for \( y \) in terms of \( x \): \[ y = 3x + 4 \]
• Substitute \( y = 3x + 4 \) into the first equation: \[ x^2 + (3x + 4)^2 = 34 \]
• Expand and simplify: \[ x^2 + (9x^2 + 24x + 16) = 34 \] \[ 10x^2 + 24x + 16 = 34 \] \[ 10x^2 + 24x - 18 = 0 \]
• Simplify by dividing through by 2: \[ 5x^2 + 12x - 9 = 0 \]
• Solve the quadratic equation using the quadratic formula: \[ x = \frac{-12 \pm \sqrt{(12)^2 - 4(5)(-9)}}{2(5)} \] \[ x = \frac{-12 \pm \sqrt{144 + 180}}{10} \] \[ x = \frac{-12 \pm \sqrt{324}}{10} \] \[ x = \frac{-12 \pm 18}{10} \]
• Thus, the two solutions for \( x \) are:
  • For the positive root: \[ x = \frac{-12 + 18}{10} = \frac{6}{10} = 0.6 \]
  • For the negative root: \[ x = \frac{-12 - 18}{10} = \frac{-30}{10} = -3 \]
• For \( x = 0.6 \), substitute into \( y = 3x + 4 \): \[ y = 3(0.6) + 4 = 1.8 + 4 = 5.8 \]
• For \( x = -3 \), substitute into \( y = 3x + 4 \): \[ y = 3(-3) + 4 = -9 + 4 = -5 \]
• Thus, the solutions are: \[ (x, y) = (0.6, 5.8) \quad \text{or} \quad (-3, -5) \]