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12. QUADRATIC EQUATIONS
There are three ways to solve quadratic equations:
(i) factorising
(ii) completing the square
(iii)
by substituting in the
formula; x = ![]()
To illustrate this we will solve the equation x2 + 2x -8 = 0 by each of the three methods.
(i) factorising: x2 + 2x -8 = 0 can be factorised to (x – 2)( x + 4) = 0
Hence (x – 2) = 0 giving x = 2 OR (x + 4) = 0 giving x = -4
x = 2 or -4
(ii) completing the square: Complete the square by adding the square of half the coefficient of x.
x2 + 2x -8 = 0 The coefficient of x is 2. Half the coefficient of x is 1. So we add 1 and subtract 1.
x2 + 2x +1 -1 -8 = 0 x2 + 2x +1 -9 = 0 (x + 1)2 – 9 = 0 (x + 1)2 = 9
taking the square root of each side: (x + 1) =
3
(x + 1) = 3 gives x = 2 (x + 1) = -3 gives x = -4
x = 2 or -4.
(iii)
The formula; x =
x2
+ 2x – 8 = 0
In this case a = 1, b = 2, c = -8
Substituting: x = ![]()
x =
= ![]()
x =
= 2 OR x
=
= -4
x = 2 or -4
Questions:
Solve the following quadratic equations by each of the above methods.
(i) x2 -2x – 8 = 0 (ii) x2 + 6x -16 = 0 (iii) x2 + 6x + 8 = 0
(iv) x2 -10x + 24 = 0 (v) x2 -12x + 36 = 0 (vi) x2 – 18x + 80 = 0
(vii) x2 – x – 2 = 0 (viii) x2 + 3x + 2 = 0 (ix) x2 + x – 6 = 0
(x) x2 + x – 12 = 0 (xi) x2 – 4x – 12 = 0 (xii) x2 + x – 20 = 0
(xiii) g2 -8g + 12 = 0 (xiv) m2 + 2m – 15 = 0 (xv) 2k2 + 7k + 3 = 0
Answers:
(i) x = 4 or -2 (ii) x = 2 or -8 (iii) x = -2 or -4
(iv) x = 4 or 6 (v) x = 6 (vi) x = 8 or 10
(vii) x = 2 or -1 (viii) x = -1 or -2 (ix) x = 2 or -3
(x) x = 3 or -4 (xi) x = 6 or -2 (xii) x = 4 or -5
(xiii) g = 2 or 6 (xiv) m = 3 or -5 (xv) k = - ½ or -3
MORE QUADRATIC
EQUATIONS
(i) x2 – 4x – 12 = 0 (ii) x2 + 7x + 12 = 0 (iii) x2 – 11x + 24 = 0
(iv) x2 + 3x - 40 = 0 (v) x2 - 10x + 16 = 0 (vi) 2x2 – x – 15 = 0
(vii) 3x2 + x – 10 = 0 (viii) 4x2 – 4x – 3 = 0 (ix) 9x2 - 6x - 8 = 0
(x) 6x2 – 13x + 6 = 0
2. Use the formula x =
to solve the following
quadratic equations. Give your answers to 1 decimal point.
(i) x2 + x - 4 = 0 (ii) x2 + 12x + 3 = 0 (iii) x2 – 2x - 5 = 0
(iv) 2x2 –10 x + 1 = 0 (v) 5x2 – 2x - 3 = 0 (vi) 2x2 + x -1 = 0
(vii) 3x2 + 6x -2 = 0 (viii) 6x2 + 4x -3 = 0 (ix) 4x2 – x – 7 = 0
(x) 3x2 + 4x – 9 = 0
3. Use the formula from Q.2. to show that the equation x2 + x + 4 = 0 has no solution.
Draw the graph y = x2 + x + 4 and state why the graph shows that the equation x2 + x + 4 = 0 has no solution.
4. Draw the graph: y = x2 – 2x – 3
Use your graph to solve the equations x2 – 2x – 3 = 0 and x2 – 2x – 8 = 0.
Answers:
1. (i) x = 6 or -2 (ii) x = -3 or -4 (iii) x = 3 or 8 (iv) x = 5 or -8 (v) x = 2 or 8
(vi) x = 3 or -2.5 (vii) x = 5/3 or -2 (viii) x = 1.5 or – 0.5 (ix) x = 4/3 or -2/3
(x) x = 1½ or 2/3
2. (i) x = 1.6 or -2.6 (ii) x = -0.3 or -11.7 (iii) x = 1.4 or 3.4 (iv) x = 0.1 or 4.9
(v) x = 1 or -0.6 (vi) x = ½ or -1 (vii) x = 0.3 or -2.3 (viii) x = 0.4 or –1.1 (ix) x = 1.5 or -1.2 (x) x = 1.2 or – 2.5
3. x =
=
graph:
Since
has no solution and
the graph does not cut the x-axis then the equation x2 + x + 4 = 0
has no solution.
4. The solutions to x2 – 2x – 3 = 0 are the points where y = 0 i.e. x = -1 or 3.
Graph:

The solution to x2 – 2x – 8 = 0 is the same as x2 – 2x – 3 = 5 i.e. the intersection of the graph y = x2 – 2x – 3 and the line y = 5 i.e. x = 4 or -2
QUADRATIC FUNCTIONS
A quadratic function is one of the form y = ax2 + bx + c where a, b &c are constants.
A graph of this function is a parabola.
If a is positive (a>0) the graph is concave up.
If a is negative (a<0) the graph is concave down.
If a = 0 the function is not quadratic and the graph is not a parabola.
The graph of a parabola may cut the x-axis at two points, touch the x-axis at one point only or lie completely above or below the x-axis.
The x values where the function touches or cuts the x-axis are known as the roots of the function.