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.

1. Example 1: Solve 3x2 + 10 = 7x + 16 .
Re-arranging to obtain: 3x2 - 7x - 6 = 0 ; factorize
(3x + 2)(x - 3)= 0
⇒ 3x + 2 = 0 or x - 3 = 0
Solutions: x= -2/3 or x = 3 .

Solve: 3x2 = 17x - 10.

2. Example 2: Solve 4x4 = 73x2- 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, (ax2)2 + (bx2) + c = 0. We can now proceed with factorization.
(4x2 - 9)(x2 - 16) = 0 ; Note both brackets are of the form (a2 - b2) 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: 4x4 = 13x2- 9.

3. Example 3: Solve (2)32x+1 = 2 - (4)3x.
Re-arrange and re-write as 2(3)(3)2x + (4)3x - 2 = 0
Let y = 3x
6y2 + 4y - 2 = 0 ; note: 32x = (3x)2 = y2.
(3y - 1)(2y + 2) = 0
⇒ y = 1/3 or y = -1
⇒ 3x = 1/3 or 3x = -1 (rejected as impossible)
⇒ x = -1 (by inspection or otherwise)

Solve: 32x+1 + (11)3x = 4.

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
⇒ x2 - 3x -5x + 15 - 3 = 0
⇒ x2 - 8x + 12 = 0
⇒ (x - 6)(x - 2) = 0
⇒ x = 6 ; (x = 2 is rejected because if x = 2 then log(2-3) is undefined)

Solve: log(x - 5) + log(x - 14) = 1

Completing the Square.

1. 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.

2. Before we go any further, this is a good time to review the special form (a - b)2.
(a - b)2 = (a2 - 2ab + b2) ; achieved through simple multiplication or binomial theorem.

3. The task is to transform any ax2 + bx + c = 0 into a(x - h)2 + k = 0. This task is "completing the square."

4. Let us start by working through an example. Solve x2 - 6x = 10.
x2 - 6x - 10 = 0 ; we need to form a perfect square here. We need to get part of the expression into (a2 - 2ab + b2). 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 32 in there. So we will add 32 - 32 to the original expression. Note that the 3 is half of the coefficient of x and the order is addition follows by deduction. Thus,

⇒ x2 -2(3)x +(32) -(32) - 10 = 0
⇒[ x2 -2(3)x +(32) ] - 9 -10 = 0 ; employ property (a - b)2 = (a2 - 2ab + b2)
⇒ (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."

 Can we always transform ax2 + bx + c = 0 into a(x - h)2 + k = 0? The answer is yes but some of the solutions may not be real numbers. a(x - h)2 + k = 0 is also called the Turning Point Form Given ax2 + bx + c = 0 , we can rewrite as a [ x2 + (b/a)x ] + c = 0 , Here we need to rearrange the thing in the bracket and complete the square. Always treat "a" as the Alien that has to be factored out from your equation. ⇒ a {x2 - 2[-b/(2a)]x + [-b/(2a)]2 - [-b/(2a)]2} + c = 0⇒ a {x2 - 2[-b/(2a)]x + [-b/(2a)]2} + { c - [-b/(2a)]2 } = 0⇒ a {x - [-b/(2a)] }2 + { c - [-b/(2a)]2 } = 0⇒ a[ x - h]2 + k where h = -b/(2a) and k = c - [-b/(2a)]2. If x= h then y= k. This point (h,k) is either the minimum or the maximum point in the curve, thus the name turning point form. We will come back to this latter.
1. Another example. Solve x2 - (5/3)x - (16/36) = 0.
⇒ x2 - 2(5/6)x + (5/6)2 - (5/6)2 - (16/36) = 0
⇒ [x2 - 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.)

2. Write the following quadratics in the turning point form using "completing the square" method

1. y = x2 + 2x

2. y = x2 - 3x + 6

3. y = 2x2 - 4x + 5

4. y = -3x2 - x + 3

5. y = (-1/2)x2 + x + (1/2)

Answers:(a) y=(x+1)2-1; (b) y=(x-3/2)2-(15/4); (c) y=2(x-1)2+3; (d) y=-3(x+1/6)2+ (37/12); (e) y= -1/2(x-1)2+ 1.

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.]

Let ax2 + bx + c = 0
then x2 + (b/a)x + (c/a) = 0
x2 - 2[-b/(2a)]x + [-b/(2a)]2 - [-b/(2a)]2} + (c/a) = 0
{x2 - 2[-b/(2a)]x + [-b/(2a)]2}+ {(c/a) - [-b/(2a)]2} = 0
(x + b/(2a) )2 = - (c/a)+ [b2/4a2]
x + b/(2a) = ± √ { [b2- 4ac] / (4a2) }
x = - [b/(2a)] ±√ { [b2- 4ac] }/ (2a)
x = {-b ± √(b2- 4ac) }/ (2a) The Discriminant.
A close inspection of the quadratic formula above will reveal that the nature of our solution can be determined by b2 - 4 ac. The expression b2 - 4ac is known as the "discriminant" and is often denoted by Δ. There are three cases to consider.

1. 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. 2. 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.

3. 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 x-axis. Please refer to the side diagram.

4. Given that y1= x2 - 12x + 28 , y2 = x2 - 2x + 8, and y3 = x2+ 5. Determine the nature of the roots of the equations above.

5. 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.

6. If a < 0 then we have an "unhappy curve" or a "frowning curve" that looks like ∩ .

7. Both "happy" and "unhappy" curves are called parabola opens upwards and parabola opens downwards respectively.

8. The maximum or the minimum point on the parabola is called the vertex. The vertex can be easily found through the "turning-point" form. If a[ x - h]2 + k , and a > 0 then as long as x ≠ h we will have (x-h)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?

9. 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.

10. Do we always have to derive the turning-point 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) = x2 - 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 y-coordinate is f(2/3) = (2/3)2 - 3(3/2)+ 5.
y-coordinate = 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) = -2x2 + 4x + 3 with -2 ≤ x ≤ 3.
Find the axis of symmetry and coordinates of the vertex. Also identify the y-intercept of f.
Solutions.
The axis of symmetry is x = -b/2a
x = -(4)/ [ 2(-2) ]
x = -4/(-4)
x = 1.
The y-ccordinate of the vertex is f(1) = -2(1)2+4(1)+3
f(1) = 5
The coordinates of the vertex are (1,5).
The y-intercept of f is when x = 0. So y intercept = -2(0)2+4(0)+3 .
The y-intercept is 3.
Note that the y-intercept is not written as (0,3).Note that the domain is not used in this question at all.

### The factor form of a quadratic function: y = a(x-h)(x-k) .

Written as y = a(x-h)(x-k) the h and k are the x-intercepts (zeroes or roots) of the function y.
The y-intercept is when x = 0 and thus y-intercept is y(0). The value of y(0) = a(0-h)(0-k).
Thus, y-intercept is ahk.

Example 7.
Find the y-intercept and roots of y = -3(x +2)(x -1).
Solutions.
The roots are x= -2 and x =1. The y-intercept is -3(-2)(1) = 6.

 With GDC. If you plot the function f with y1=-3(x+2)(x-1) then to obtain the y-intecept you simply use [2nd][TRACE]select 1:value[ENTER] and at the prompt x={blinking} type in [ENTER]. You will see the y-coordinate given as 6. To find the roots it is easier to plot y2=0 and then use intersection to find the roots. Personally, I will prefer to memorize the formula above as it is easier.

Example 8. Find the y-intercept 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 x-axis only once.
The y-intercept 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(x-h)(x-k). y = a(x-(-1))(x-3) y = a(x +1)(x -3) But we also know that the y-intercept 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)(x-3). (b) Solution: We will use the form y = a(x-h)(x-k). y = a(x-(-0.5))(x-1.5) y =a(x+0.5)(x-1.5) And the y-intercept is 1.5 = a(-0.5)(1.5) a= 1/(-0.5) a=-2 So the quadratic equation is y = -2(x+0.5)(x-1.5) or y = -2(x+1/2)(x-3/2)

Questions:

1. [Calculator is allowed]
The height of a projectile from its origin could be modelled using h = -2x2+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(x-3/2)(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 (x-5)(x+1) = -6x, show that it has two real solutions.
(b) Without using a calculator, solve (x-5)(x+1) = -6x.

4. [Calculator is allowed]
Given that a quadratic function f has a y-intercept at -7.5 and intersects the x-axis at x = 1.5 and x = -1.