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Study Guide For Grade

11

maths

Term 4
Paper 1

Algebraic Expressions, Equations and Inequalities

Algebraic expressions contain variables, constants and operations. An equation states that two expressions are equal, while an inequality compares expressions. In examinations, show all algebraic steps, state restrictions where necessary and check solutions that may have been introduced by squaring.

Exponents and surds

An exponent shows how many times a base is used as a factor. Negative exponents represent reciprocals, while fractional exponents represent roots.

an=1an
a1n=an
amn=amn

A surd is an irrational root that cannot be simplified to a rational number. Examples include:

2,5,37

A square root is defined only when its radicand is non-negative in the real number system.

Laws of exponents

For non-zero bases, apply the following laws:

am×an=am+n
aman=amn
(am)n=amn
(ab)n=anbn
(ab)n=anbn
a0=1

Example

Simplify the expression.

6x5y23x2y4
=2x52y2(4)
=2x3y2

Common Mistake

Do not apply exponent laws to addition. In general:

am+anam+n

Simplifying expressions involving surds

To simplify a surd, identify a perfect-square factor.

Example

Simplify the surd.

72
=36×2
=62

Only like surds may be added or subtracted.

35+25=55

To remove a surd from a denominator, rationalise the denominator.

Example

43
=43×33
=433

For a denominator containing two terms, multiply by the conjugate.

(a+b)(ab)=a2b2
12+3×2323
=2343
=23

Equations involving surds

Isolate the surd before squaring both sides. Squaring can introduce extraneous solutions, so substitute answers into the original equation.

Example

Solve the equation.

2x+3=x

Since a square root is non-negative:

x0

Square both sides.

2x+3=x2
x22x3=0
(x3)(x+1)=0
x=3orx=1

Check the possible solutions. The value negative one does not satisfy the original equation. Therefore:

x=3

Quadratic equations

A quadratic equation has the standard form:

ax2+bx+c=0,a0

Quadratic equations can be solved by factorisation, completing the square or using the quadratic formula. Always write the equation in standard form before choosing a method.

The solutions are also called roots or zeros. Graphically, real roots are the x-coordinates of the x-intercepts of the corresponding parabola.

Factorisation

Factorisation rewrites a quadratic expression as a product of two factors. Use the zero-product property:

AB=0A=0orB=0

Example

Solve by factorisation.

2x2x6=0

Split the middle term.

2x2+3x4x6=0
x(2x+3)2(2x+3)=0
(2x+3)(x2)=0
2x+3=0orx2=0
x=32orx=2

Remember to look for a highest common factor before using other factorisation methods.

Completing the square

Completing the square changes a quadratic expression into the form:

a(xp)2+q

Example

Solve by completing the square.

x26x+5=0

Move the constant term.

x26x=5

Add the square of half the coefficient of the linear term to both sides.

x26x+9=5+9
(x3)2=4
x3=±2
x=3±2
x=1orx=5

Exam Tip

When the coefficient of the squared term is not one, first divide every term by that coefficient or factor it out.

Quadratic formula

The quadratic formula solves any quadratic equation in standard form.

x=b±b24ac2a

Example

Solve using the quadratic formula.

2x2+3x1=0

Identify the coefficients.

a=2,b=3,c=1

Substitute carefully.

x=3±324(2)(1)2(2)
x=3±9+84
x=3±174

Leave exact answers in surd form unless a decimal approximation is requested.

Quadratic inequalities

A quadratic inequality may contain signs such as less than, greater than, less than or equal to, or greater than or equal to. First find the critical values by solving the related quadratic equation. Then use a sign table or a sketch of the parabola.

Example

Solve the inequality.

x2x60

Factorise.

(x3)(x+2)0

Find the critical values.

x=2orx=3

The parabola opens upwards, so the expression is non-positive between the roots. The equality sign means the endpoints are included.

2x3

If the inequality is strict, the endpoints are excluded.

Simultaneous equations

Simultaneous equations are equations that must be satisfied by the same values of the variables. Use substitution or elimination.

Example

Solve the system.

y=x+1
x2+y=7

Substitute the first equation into the second.

x2+x+1=7
x2+x6=0
(x+3)(x2)=0
x=3orx=2

Calculate the corresponding values of the second variable.

y=3+1=2
y=2+1=3

The solutions are:

(x,y)=(3,2)or(x,y)=(2,3)

On the Cartesian plane, these solutions represent the points of intersection of the graphs.

Linear inequalities

Solve a linear inequality in the same way as a linear equation. However, reverse the inequality sign when multiplying or dividing by a negative number.

Example

3x5<10
3x<15
x<5

Example involving a negative coefficient:

2x+311
2x8

Divide by negative two and reverse the inequality sign.

x4

For a compound inequality, perform the same operation on all three parts.

3<2x+17
4<2x6
2<x3

Nature of roots

The nature of the roots describes the number and type of solutions of a quadratic equation. A quadratic can have:

  • Two distinct real roots.
  • Two equal real roots.
  • No real roots.

If the roots are not real, they are non-real. The nature of the roots can be determined without solving the equation completely.

Discriminant

The discriminant is the expression inside the square root in the quadratic formula.

Δ=b24ac

The value of the discriminant determines the nature of the roots.

Δ>0

There are two distinct real roots. If the discriminant is also a perfect square, the roots are rational; otherwise, they are irrational.

Δ=0

There are two equal real roots.

Δ<0

There are no real roots.

Example

Determine the nature of the roots.

3x24x+2=0
a=3,b=4,c=2
Δ=(4)24(3)(2)
Δ=1624
Δ=8

Since the discriminant is negative, the equation has no real roots.

Word problems involving equations and inequalities

Translate words into algebra before solving. Define the unknown, form an equation or inequality, solve it and check whether the answer is reasonable in the context.

Example

The length of a rectangle is three metres more than its width. Its area is forty square metres. Calculate its dimensions.

Let the width be represented by a variable. Then the length is three more than the width.

x(x+3)=40
x2+3x40=0
(x+8)(x5)=0
x=8orx=5

A measurement cannot be negative, so the width is five metres.

x+3=8

The rectangle is five metres wide and eight metres long.

Example

A learner must score at least sixty marks over two tests. The learner obtained twenty-seven marks in the first test. Determine the minimum mark required in the second test.

Let the second-test mark be represented by a variable.

27+x60
x33

The learner must obtain at least thirty-three marks.

Exam Tip

In word problems, reject solutions that do not satisfy practical restrictions. Ages, lengths, quantities and time values are normally non-negative. For inequalities, words such as “at least” indicate greater than or equal to, while “at most” indicate less than or equal to.