Aims: By the end of this section, you will be able to
(a) solve equations that are in quadratic form using factorization or quadratic
formula,
(b) complete the square,
(c) use properties associated with the turning point form,
(d) calculate and use discriminant of a quadratic function, and
(e) manipulate graphical properties of a quadratic function.
Quadratic Form Factorization.
Example 1: Solve 3x^{2}
+ 10 = 7x + 16 .
Rearranging to obtain: 3x^{2} 
7x  6 = 0 ; factorize
(3x + 2)(x  3)= 0
⇒ 3x + 2
= 0 or x  3 = 0
Solutions: x= 2/3 or x = 3 .
Solve:
3x^{2} = 17x  10.
Example 2: Solve 4x^{4} = 73x^{2}
144.
This is an expression in the 4th degree polynomial
but if you study it carefully then you will realize that it has a
quadratic form. Namely, (ax^{2})^{2} + (bx^{2})
+ c = 0. We can now proceed with factorization.
(4x^{2}
 9)(x^{2}  16) = 0 ; Note both brackets are of the form
(a^{2}  b^{2}) and we can easily factorize
further.
(2x + 3)(2x  3)(x + 4)(x  4) = 0.
⇒ x =
3/2 or x = 3/2 or x = 4 or x =4.
Solve:
4x^{4} = 13x^{2} 9.
Example 3: Solve (2)3^{2x+1} =
2  (4)3^{x}.
Rearrange and rewrite as
2(3)(3)^{2x} + (4)3^{x}  2 = 0
Let y = 3^{x}
6y^{2} + 4y  2 = 0 ; note: 3^{2x} = (3^{x})^{2}
= y^{2}.
(3y  1)(2y + 2) = 0
⇒ y = 1/3 or y =
1
⇒ 3^{x} = 1/3 or 3^{x} = 1 (rejected as
impossible)
⇒ x = 1 (by inspection or otherwise)
Solve:
3^{2x+1} + (11)3^{x} = 4.
Example 4: Solve log(x  5) + log(x 
3) = log (3).
⇒ log[ (x  5)(x  3) ] = log (3);
logarithmic law is used
⇒ (x  5)(x  3) = 3
⇒
x^{2}  3x 5x + 15  3 = 0
⇒ x^{2}  8x + 12
= 0
⇒ (x  6)(x  2) = 0
⇒ x = 6 ; (x = 2 is
rejected because if x = 2 then log(23) is undefined)
Solve:
log(x  5) + log(x  14) = 1
Completing the Square.
If we have (x  a)^{2}
= b then
(x  a) = ± √(b)
⇒ x = a + √(b) or x = a  √(b)
This solution method requires a perfect square on one side and a number
on the other side. We use "completing the square" method to obtain
this desirable condition.
Before we go any further, this is a good time
to review the special form (a  b)^{2}.
(a  b)^{2} = (a^{2}  2ab + b^{2})
; achieved through simple multiplication or binomial theorem.
The task is to transform any ax^{2} + bx + c = 0 into a(x  h)^{2} + k = 0. This task is "completing the square."
Let us start by working through an example. Solve x^{2}  6x =
10.
x^{2}  6x  10 = 0 ; we need to form a perfect square here.
We need to get part of the expression into (a^{2}  2ab + b^{2}).
The middle term is negative and twice the product of a and b. The last term
is the square of b. Rewriting 6x as 2(3)x quickly reveals to us the value
of b which is 3 in this case. Somehow, we need to have 3^{2} in
there. So we will add 3^{2}  3^{2} to
the original expression. Note that the 3 is half of the coefficient of
x and the order is addition follows by deduction. Thus,
⇒ x^{2} 2(3)x +(3^{2})
(3^{2})  10 = 0
⇒[ x^{2} 2(3)x +(3^{2})
]  9 10 = 0 ; employ property (a  b)^{2} = (a^{2}
 2ab + b^{2})
⇒ (x  3)^{2} = 19
⇒ x  3 = √(19)
⇒ x = 3 + √(19) or x = 3  √(19).
⇒ x ≈ 7.36 or x ≈ 1.36 (3.s.f)
Key : "To complete the square we add half of the coefficient of x into and deduct half of the coefficient of x from the equation."

Another example. Solve x^{2}  (5/3)x
 (16/36) = 0.
⇒ x^{2}  2(5/6)x + (5/6)^{2}
 (5/6)^{2}  (16/36) = 0
⇒ [x^{2}  2(5/6) x + (5/6)^{2}]  25/36 16/36 = 0
⇒ (x  (5/6))^{2} = 41/36
⇒ x = 5/6 + √(41/36) or x = 5/6  √(41/36)
⇒ x ≈ 1.90 or x ≈ 0.234 (3.s.f.)
Write the following quadratics in the turning point form using "completing the square" method
y = x^{2} + 2x
y = x^{2}  3x + 6
y = 2x^{2}  4x + 5
y = 3x^{2}  x + 3
y = (1/2)x^{2} + x + (1/2)
Answers:(a) y=(x+1)^{2}1; (b) y=(x3/2)^{2}(15/4); (c) y=2(x1)^{2}+3; (d) y=3(x+1/6)^{2}+ (37/12); (e) y= 1/2(x1)^{2}+ 1.
Quadratic Formula.
Obviously, the completing the square
method could be a bit tedious and it would be easier if we can plug
numbers into a formula for any question that says "solve
for x" or "find the roots of f(x)"
or even "the zeroes of a function." We
achieve this by using the quadratic formula. [
Refer to the previous note on factorization if you need help on using
quadratic formula to solve quadratic equations.]
Deriving
the quadratic formula.
Let ax^{2} + bx + c = 0
then x^{2} + (b/a)x + (c/a) = 0
x^{2} 
2[b/(2a)]x + [b/(2a)]^{2}  [b/(2a)]^{2}} + (c/a)
= 0
{x^{2}  2[b/(2a)]x + [b/(2a)]^{2}}+ {(c/a)
 [b/(2a)]^{2}} = 0
(x + b/(2a) )^{2} =  (c/a)+
[b^{2}/4a^{2}]
x + b/(2a) =
± √ { [b^{2}
4ac] / (4a^{2}) }
x =  [b/(2a)]
±√ { [b^{2}
4ac] }/ (2a)
x = {b ± √(b^{2}
4ac) }/ (2a)
The Discriminant.
A close inspection of the quadratic
formula above will reveal that the nature of our solution can be
determined by b^{2}  4 ac. The expression b^{2} 
4ac is known as the "discriminant" and is often denoted by
Δ. There are three cases to consider.
Case 1: Δ = 0 .
If Δ = 0 then √(0) is 0 then x = b/2a. Thus, there is only
one root and the root is a real number.
Case 2: Δ > 0 .
If Δ > 0 then √(Δ) is either a positive real number
or a negative real number. Thus, there are two roots and both are real numbers.
Case 3: Δ < 0 .
If Δ < 0 then √(Δ) is NOT a real number. Thus, there
is no real number root. The curve does not intersects with the xaxis. Please
refer to the side diagram.
Given that y1= x^{2}  12x + 28 , y2 = x^{2}  2x + 8, and y3 = x^{2}+ 5. Determine the nature of the roots of the equations above.
Also note from the diagram at the side that if a > 0 then we have a "happy curve" or a "smilling curve". The shape of the curve would be similar to È.
If a < 0 then we have an "unhappy curve" or a "frowning curve" that looks like ∩ .
Both "happy" and "unhappy" curves are called parabola opens upwards and parabola opens downwards respectively.
The maximum or the minimum point on the parabola is called the vertex. The vertex can be easily found through the "turningpoint" form. If a[ x  h]^{2} + k , and a > 0 then as long as x ≠ h we will have (xh)^{2} > 0. Thus, the total value will be greater than k. If x = h then we will obtain the minimum for the function, namely y = k. This means, (h, k) is the minimum point. Could you tell similar story to justify (h, k) for being the maximum point when a < 0?
If you study all the curves above then you will find that these curves are symmetric. The line of symmetry actually runs through the maximum/minimum point. Thus, the line of symmetry is simply x = h.
Do we always have to derive the turningpoint form to obtain the maximum/minimum
point and the line of symmetry? The answer is NO. Look at the derivation
in the box above. Notice that h= b/(2a).
So the maximum/minimum point is (b/(2a), y(b/(2a)) and the line of symmetry
is x = b/(2a).
Example 5: Given that f(x) = x^{2}  3x +5 .
(a) Find the axis of symmetry of the function f.
(b) Identify the nature of the vertex in function f.
(b) Find the coordinates of this vertex.
Solutions.
(a) The axis of symmetry is x = b/2a
The axis of symmetry is x = (3)/2(1)  Notice, it is a good practice
to use a bracket for negative number.
The axis of symmetry is x = 3/2
(b) Since the a=1 > 0 this is a parabola that opens upward. Thus, the vertex
is a local minimum point.
(c) x coordinate is 3/2 as in (a). The ycoordinate is f(2/3) = (2/3)^{2}
 3(3/2)+ 5.
ycoordinate = 11/4 or 2.75.
The coordinates of the vertex are (3/2, 11/4)
You could also use your GDC to visualize this by plotting
the graph for function f. But in this case it is faster to use the formula.
Example 6: Given that f (x) = 2x^{2} + 4x + 3 with
2 ≤ x ≤ 3.
Find the axis of symmetry and coordinates of the vertex. Also identify the yintercept
of f.
Solutions.
The axis of symmetry is x = b/2a
x = (4)/ [ 2(2) ]
x = 4/(4)
x = 1.
The yccordinate of the vertex is f(1) = 2(1)^{2}+4(1)+3
f(1) = 5
The coordinates of the vertex are (1,5).
The yintercept of f is when x = 0. So y intercept = 2(0)^{2}+4(0)+3
.
The yintercept is 3.
Note that the yintercept is not written as (0,3).Note that the domain is not
used in this question at all.
Written as y = a(xh)(xk) the h and k are the xintercepts
(zeroes or roots) of the function y.
The yintercept is when x = 0 and thus yintercept is y(0). The value of y(0)
= a(0h)(0k).
Thus, yintercept is ahk.
Example 7. Find the yintercept and roots of y = 3(x
+2)(x 1).
Solutions.
The roots are x= 2 and x =1. The yintercept is 3(2)(1) = 6.
With GDC. Personally, I will prefer to memorize the formula above as it is easier. 
Example 8. Find the yintercept and roots of y = (x+4)^{2}/6
.
Solutions.
The function can be rewritten as y = (1/6)(x+4)(x+4).
Basically the two roots are not distinct. x = 4. That is the
graph intersects the xaxis only once.
The yintercept is (1/6)(4)(4) = 8/3
Example 9. Find the quadratic equation for the given graphs.
(a) Solution: We will use the form y = a(xh)(xk). y = a(x(1))(x3) y = a(x +1)(x 3) But we also know that the yintercept is 4.5 thus 4.5 = a(1)(3) a = 4.5/(3) a = 3/2. So the quadratic equation is y = (3/2)(x+1)(x3). 
(b) Solution: We will use the form y = a(xh)(xk). y = a(x(0.5))(x1.5) y =a(x+0.5)(x1.5) And the yintercept is 1.5 = a(0.5)(1.5) a= 1/(0.5) a=2 So the quadratic equation is y = 2(x+0.5)(x1.5) or y = 2(x+1/2)(x3/2) 
Questions:
1. [Calculator is allowed]
The height of a projectile from its origin could be modelled using h = 2x^{2}+20x,
where x is the horizontal displacement from the origin at ground level.
(a) How far is it from the origin measured at ground level when the projectile
is at its maximum height?
(b) What is the maximum height of this projectile?
2. [No calculator]
(a) By completing the square, show that y = 2(x3/2)(x+1/2)5 is always negative
for all real value of x.
(b) Confirm that the discriminant for the above function in (a) is negative.
3. (a) Without using a calculator and solving this equation
(x5)(x+1) = 6x, show that it has two real solutions.
(b) Without using a calculator, solve (x5)(x+1) = 6x.
4. [Calculator is allowed]
Given that a quadratic function f has a yintercept at 7.5 and intersects the
xaxis at x = 1.5 and x = 1.
Find this quadratic function.
5. [Calculator is allowed]
The roots of a quadratic function y are x = 4 and x = 6. A point (2,48) is
on the graph y.
Find this quadratic function.
Answers: 1(a) xmax= 20/(2(2)) = 5, 1(b) hmax=h(5) =50; 2(a). show that y=2(x1/2)^{2}3, since the value a = 2 then the curve opens downward with a maximum y value of only 3. Thus, it must be the case that the function y is always negative for all real value x. 2(b) show that the dicriminant value is 24 which is negative. 3(a) show that discriminant is positive and conclude that there are two real solutions. 3(b) x = 1√6 or x =1+√6. (4) f(x) = 5(x3/2)(x+1). (5) y = 2(x6)(x+4).