Solving Quadratic Equations & Sketching Parabolas

Solving Quadratic Equations

  • Quadratic equations often need to be solved when sketching graphs or dealing with quadratic relationships.
  • Solutions can be found by plotting a quadratic graph and finding x-intercepts, but this can be slow and inaccurate.
  • Algebraic techniques provide exact solutions.
  • Not all quadratic equations have real solutions (i.e., they don't all cross the x-axis).

Null Factor Law

  • Factorise the equation and use the null factor law to create a linear equation.
  • The null factor law states: If ab = 0, then a = 0 or b = 0 or a = b = 0.

Worked Example 1

Solve for x:

(a) x(x - 2) = 0

  • Equate each factor to zero: x = 0 or x - 2 = 0
  • Solve: x = 0 or x = 2

(b) (x + 1)(x - 6) = 0

  • Equate each factor to zero: x + 1 = 0 or x - 6 = 0
  • Solve: x = -1 or x = 6

Worked Example 2

Solve for x:

(a) 4x^2 = 0

  • Divide both sides by 4: x^2 = 0
  • Equate the linear factor to zero: x = 0

(b) (x + 1)^2 = 0

  • Equate the linear factor to zero: x + 1 = 0
  • Solve: x = -1

Factorisation Techniques

  • Before using the null factor law, ensure the quadratic equation is factorised.
  • Techniques include:
    • Taking out common factors (including a negative sign if appropriate).
    • Using the difference of two squares rule.
    • Using perfect square rules.
    • Factorising monic trinomials.
  • After factorising, apply the null factor law.
  • Quadratics that cannot be factorised using real numbers have no real solutions.
  • Perfect square quadratics have one solution.
  • Otherwise, there will be two solutions.

Worked Example 3

Solve for x:

(a) 3x^2 - 6x = 0

  • Factorise: 3x(x - 2) = 0
  • Apply the null factor law: 3x = 0 or x - 2 = 0
  • Solve: x = 0 or x = 2

(b) 11x^2 = x

  • Rearrange: 11x^2 - x = 0
  • Factorise: x(11x - 1) = 0
  • Apply the null factor law: x = 0 or 11x - 1 = 0
  • Solve: x = 0 or x = \frac{1}{11}

Worked Example 6

Solve for x:

(a) x^2 - 6x + 8 = 0

  • Factorise: (x - 2)(x - 4) = 0
  • Apply the null factor law: x - 2 = 0 or x - 4 = 0
  • Solve: x = 2 or x = 4

(b) 2x^2 + 8x = 24

  • Rearrange: 2x^2 + 8x - 24 = 0
  • Factorise: 2(x^2 + 4x - 12) = 0
  • Divide by 2: x^2 + 4x - 12 = 0
  • Factorise: (x + 6)(x - 2) = 0
  • Apply the null factor law: x + 6 = 0 or x - 2 = 0
  • Solve: x = -6 or x = 2

(c) -x^2 - 4x - 3 = 0

  • Factor out -1: -1(x^2 + 4x + 3) = 0
  • Divide by -1: x^2 + 4x + 3 = 0
  • Factorise: (x + 1)(x + 3) = 0
  • Apply the null factor law: x + 1 = 0 or x + 3 = 0
  • Solve: x = -1 or x = -3

Methods for Solving Quadratic Equations

  1. Equate one side to zero.
  2. Factorise by looking for:
    • A common factor
    • A difference of two squares: a^2 - b^2 = (a + b)(a - b)
    • A perfect square: a^2 \pm 2ab + b^2 = (a \pm b)^2
    • A monic trinomial (use the cross method or splitting the middle term).
    • You may need to take out the coefficient of x^2 as a common factor and divide both sides by the common factor before factorising.

Factorising by Completing the Square

Factorising Four Terms by Grouping 'Three and One'

  • Not all expressions with four terms can be factorised by grouping in pairs.
  • An expression with three terms and one term can be factorised if the three terms form a perfect square and the fourth term is subtracted to form a difference of two squares.
    • Identify a perfect square and group the three terms.
    • Factorise the three terms using a perfect square rule.
    • Write the expression as a difference of two squares.
    • Use the difference of two squares rule to complete the factorisation.

Worked Example 7

Factorise x^2 + 4x + 4 - y^2 by grouping 'three and one'.

  • Identify a perfect square and group the three terms: (x^2 + 4x + 4) - y^2
  • Factorise using the perfect square rule: (x + 2)^2 - y^2
  • Use the difference of two squares rule: (x + 2 - y)(x + 2 + y)

Completing the Square

  • If you can't find two numbers that multiply to give c and add to give b in the quadratic trinomial x^2 + bx + c, you can use the completing the square method of factorising.
    • Add a third term to the first two terms to make a perfect square.
    • Subtract the amount added to keep the expression equivalent.
    • Identify a 'three and one' expression.
    • Factorise using a perfect square rule and the difference of two squares rule.

Solving by Completing the Square

  • Use this method to solve quadratic equations that cannot be solved easily using other methods.
    • Equate the expression to zero.
    • Look for a common factor.
    • Add and subtract terms in the expression to make a perfect square.
    • Factorise using the difference of two squares rule.
    • Solve using the null factor law.

Worked Example 10

For each of the following expressions:

(a) x^2 - 6x + 3

(i) Complete the square and factorise.

  • Write the expression. Use the first two terms to create a perfect square. The term that is added and subtracted is the square of half the coefficient of the x term.

x^2 - 6x + 3 = (x^2 - 6x + (\frac{6}{2})^2) - (\frac{6}{2})^2 + 3

  • Factorise the perfect square and simplify the constant term.

= (x^2 - 6x + (-3)^2) - 9 + 3 = (x - 3)^2 - 6

  • Rewrite the expression with the constant term as a square.

= (x - 3)^2 - (\sqrt{6})^2

  • Factorise the expression using the difference of two squares rule.

= (x - 3 + \sqrt{6})(x - 3 - \sqrt{6})

(ii) Equate the expression to zero and use your answer for part (i) to solve the quadratic equation formed.

  • Equate the expression to zero:

x^2 - 6x + 3 = 0

  • Use the factorised expression from part (i) to rewrite the equation.

(x - 3 + \sqrt{6})(x - 3 - \sqrt{6}) = 0

  • Use the null factor law.

(x - 3 + \sqrt{6}) = 0 \text{ or } (x - 3 - \sqrt{6}) = 0

  • Solve for x.

x = 3 - \sqrt{6} \text{ or } x = 3 + \sqrt{6}

  • State the solution.

x = 3 \pm \sqrt{6}

(b) x^2 + 5x - 2

(i) Complete the square and factorise.

  • Write the expression. Use the first two terms to create a perfect square. The term that is added and subtracted is the square of half the coefficient of the x term.

x^2 + 5x - 2 = x^2 + 5x + (\frac{5}{2})^2 - (\frac{5}{2})^2 - 2

  • Factorise the perfect square and simplify the constant term.

= (x + \frac{5}{2})^2 - \frac{25}{4} - 2 = (x + \frac{5}{2})^2 - \frac{33}{4}

  • Rewrite the expression with the constant term as a square.

= (x + \frac{5}{2})^2 - (\sqrt{\frac{33}{4}})^2

  • Factorise the expression using the difference of two squares rule.

= (x + \frac{5}{2} + \sqrt{\frac{33}{4}})(x + \frac{5}{2} - \sqrt{\frac{33}{4}})

(ii) Equate the expression to zero.

  • Use the factorised expression from part (i) to rewrite the equation.

(x + \frac{5}{2} + \sqrt{\frac{33}{4}})(x + \frac{5}{2} - \sqrt{\frac{33}{4}}) = 0

  • Use the null factor law.

(x + \frac{5}{2} + \sqrt{\frac{33}{4}}) = 0 \text{ or } (x + \frac{5}{2} - \sqrt{\frac{33}{4}}) = 0

  • Solve for x and simplify your answer.

x = -\frac{5}{2} - \sqrt{\frac{33}{4}} \text{ or } x = -\frac{5}{2} + \sqrt{\frac{33}{4}}

  • State the solution.

Worked Example 11

Solve x^2 + 6x - 2 = 0 using the method of completing the square. Leave your answer in exact surd form.

  • Write the equation.

x^2 + 6x - 2 = 0

  • Factorise the non-zero expression by completing the square. Use the first two terms to create a perfect square. The term that is subtracted is the square of half the coefficient of the x term.

x^2 + 6x + (\frac{6}{2})^2 - (\frac{6}{2})^2 - 2 = 0

(x^2 + 6x + 9) - 9 - 2 = 0

  • Factorise the perfect square and simplify the constant term.

(x + 3)^2 - 11 = 0

  • Factorise the expression using the difference of two squares rule.

(x + 3)^2 - (\sqrt{11})^2 = 0

(x + 3 - \sqrt{11})(x + 3 + \sqrt{11}) = 0

  • Use the null factor law.

x + 3 - \sqrt{11} = 0 \text{ or } x + 3 + \sqrt{11} = 0

  • Solve the individual equations.

x = -3 + \sqrt{11} \text{ or } x = -3 - \sqrt{11}

  • State the solution.

x = -3 \pm \sqrt{11}

Sketching Parabolas

  • A sketch graph shows the general shape and important features without completing a table of values.
  • In this section, you will sketch graphs of y = ax^2 + bx + c, where a = \pm 1.
  • To make a sketch of a quadratic graph, you will need to identify the following:

  1. The shape of the graph:

    • A positive coefficient of x^2 produces a positive parabola.
    • A negative coefficient of x^2 produces a negative parabola.

  2. The y-intercept:

    • Where the graph crosses the y-axis (when x = 0).
    • Found by substituting x = 0 into the quadratic equation.

  3. The x-intercept(s):

    • Where the graph crosses the x-axis (when y = 0).
    • There may be no x-intercepts, one x-intercept, or two x-intercepts.
    • Found by equating the quadratic to zero and solving.
      * If there is no solution, there are no x-intercepts.
      * If there is one solution, the parabola touches the x-axis. There is only one x-intercept.
      * If there are two solutions, there are two distinct x-intercepts. These may involve surds.

  4. The turning point:

    • Found by completing the square to transform the equation into turning point form.
    • The graph of y = a(x - h)^2 + k has a turning point at (h, k).
      * When a graph has two x-intercepts, the x-coordinate of the turning point is halfway between the two x-intercepts. The corresponding y-coordinate can be found by substituting the x-value into the quadratic equation.
      * When a graph has only one x-intercept, the turning point is the x-intercept.
    • The axis of symmetry is a vertical line that passes through the turning point, dividing the graph into mirror image halves.

Worked Example 12

Sketch the graph of the quadratic relationship y = x^2 + 2x - 8 by finding:

(a) the shape of the graph

  • This is a \cup-shaped parabola as the coefficient of x is positive.

(b) the y-intercept

  • x = 0, y = 0^2 + 2 \times 0 - 8
    ***y = -8
  • The y-intercept is (0, -8).

(c) the x-intercepts

  • y = 0. 0 = x^2 + 2x - 8
  • x^2 + 2x - 8 = 0
  • (x + 4)(x - 2) = 0
  • x + 4 = 0 \text{ or } x - 2 = 0
    ***x = -4 \text{ or } x = 2
  • The x-intercepts are (-4, 0) and (2, 0).

(d) the turning point

  • For the turning point:
    ***x = \frac{-4 + 2}{2} = -1
    ***x = -1, y = (-1)^2 + 2 \times -1 - 8
    ***y = 1 - 2 - 8 = -9
  • The turning point is at (-1, -9).

Alternatively, complete the square and write in turning point form.

y = x^2 + 2x - 8

y = x^2 + 2x + 1 - 8 - 1
***y = (x + 1)^2 - 9

  • The turning point is at (-1, -9).

(e) the axis of symmetry

Key features of a parabola

When sketching a parabola, the key features to show are:

  • the y-intercept
  • the x-intercepts, if they exist
  • the turning point, or vertex
  • the axis of symmetry.

Worked example 15

Sketch the graph of the equation y=-(x-2)^2+5, showing all the key features.

  1. Write the equation.

    y=-(x-2)^2+5

  2. Compare the equation to y=(x-h)^2+k to find (h, k), the turning point.

    • The graph of y=x^2 needs to be inverted and translated 2 units right and 5 units up.
    • h=2 and k=5
      The turning point is at (2,5).

  3. Find the y-intercept by substituting x=0 into the equation.

    y=-(0-2)^2+5=-(4)+5=1

    The y-intercept is (0, 1).

  4. Find the x-intercepts by substituting y=0 into the equation.

    0=-(x-2)^2+5

Solve the equation.

(x-2)^2-5=0 ((x-2)+\sqrt{5})((x-2)-\sqrt{5})=0 x=(2-\sqrt{5}) and x=(2+\sqrt{5})

x=2-\sqrt{5} or x=2+\sqrt{5}

  1. Use a calculator to find approximate values for x.

    x \approx 4.2 and -0.2