LEARNING OBJECTIVES:

• To solve the roots of a quadratic equation using
• factoring/factorising
• completing the square
• To use discriminant to determine the nature of the roots of the quadratic equation
• To describe the properties of the graph of the quadratic function

MATH CONCEPTS

• QUADRATIC EQUATION –  the general form of quadratic equation is $ax^{2}+bx+c$, where a, b, and c are constants and $a\neq 0$.
• FACTORING/FACTORISING- is the mathematical process of finding the factors of a number or expression.
• COMPLETING THE SQUARE –  is a process used to solve a quadratic equation by rewriting the form of the equation so that the left side is a perfect square trinomial.
• QUADRATIC FORMULA – is a method that is used to find the roots of a quadratic equation from its coefficients. The quadratic formula is:

$x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

• DISCRIMINANT –  the number $d= b^{2}-4ac$ determined from the coefficients of the equation $ax^{2}+bx+c$. The discriminant reveals what type of roots the equation has. It can be real or imaginary/complex roots.
• MINIMUM/MAXIMUM VALUE- The minimum value of a function is the place where the graph has a vertex at its lowest point while the maximum value of a function is the place where the graph has a vertex at its highest point.

IMPORTANCE

• Quadratic equation has many applications in everyday life. It can be used when calculating areas, determining a product’s profit or formulating the speed of an object.
• Quadratic equations are also useful in calculating speeds. Avid kayakers, for example, use quadratic equations to estimate their speed when going up and down a river.
• In athletic events that involve throwing objects like the shot put, balls or javelin, quadratic equations become highly useful.
• In the real world, you can use the minimum value of a quadratic function to determine minimum cost or area. It has practical uses in science, architecture and business.

DISCUSSION

Quadratic equation in general form is $ax^{2}+bx+c$, where a, b, and c are constants and $a\neq 0$. It is very important that the value of a should not be zero because that will make the equation linear and not quadratic anymore.

Quadratic equations come in different forms.

Note: Vertex of the parabola – it is the turning point of the graph of a quadratic equation. It contains the minimum/maximum value. It can be the peak of the graph or the lowest point.

One of the important concepts about quadratic equations is finding its roots. There are many ways to solve for the roots of the quadratic equation —- factoring/factorising, completing the square, and using quadratic formula.

Factoring/Factorising

ILLUSTRATIVE EXAMPLES

Using factorising, find the roots of the following quadratics.

1) $x^{2}+6x+8$               2) $x^{2}-5x+6$

SOLUTIONS:

1) For this given, a = 1, b = 6, c = 8. We should think of two numbers that when multiplied together, give a product of 8 and when added, they will be equal to 6. Thus, the numbers are 2 and 4. So,

$x^{2}+6x+8=(x+2)(x+4)$ Rewrite the equation in factored form

(x + 2)(x + 4) = 0  Equate the expression to 0.

$f_{1}: (x+2)=0$      $f_{2}: (x+4)=0$  Equate each factor to 0.

$r_{1}: x=-2$   $r_{2}: x=-4$

The roots of $x^{2}+6x+8$ are -2 and -4.

2) For this given, a = 1, b = -5, c = 6. We should think of two numbers that when multiplied together, give a product of 6 and when added, they will be equal to -5. Take note that the product must be positive but the sum is negative. So we need two negative numbers. Thus, the numbers are -3 and -2. So,

$x^{2}-5x+6=(x-2)(x-3)$   Rewrite the equation in factored form

(x – 2) (x – 3) = 0   Equate the expression to 0.

$f_{1}: (x-2)=0$   $f_{2}: (x-3)=0$ Equate each factor to 0.

$r_{1}: x=2$   $r_{2}: x=3$

The roots of $x^{2}-5x+6$ are 2 and 3.

Note: Not all quadratics can be factored. So completing the square method can be an alternative process.

Completing the Square

Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.

To solve $ax^{2}+bx+c$, using completing the square:

1. Rewrite the equation in the form $ax^{2}+bx=-c$. In this case, the constant will be on the right side of the equation.
2. If $a\neq 1$, divide both sides by a.
3. Divide b by 2a then square quotient $(\frac{-b}{2a})^{2}$. Add this to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Extract the square root of the both sides of the equation (Note: $(x+p)^{2}=r$ is equivalent to $x+p = \pm \sqrt{r}$.
6. Solve for the value of x.

ILLUSTRATIVE EXAMPLES

1. Find the roots using completing the square method.

1) $x^{2}-8x+1=0$           2) $2x^{2}+5x+4=0$

SOLUTIONS:

1) $x^{2}-8x+1=0$

$x^{2}-8x=-1$

$b= -8$     $-\frac{-8}{2(1)}=4$     $4^{2}=16$

$x^{2}-8x+16 =-1+16$

$x^{2}-8x+16 =15$

$(x-4)^{2} =15$

$\sqrt{(x-4)^{2}}=\sqrt{15}$

$x-4=\sqrt{15}$

$x=4\pm \sqrt{15}$

2) $2x^{2}+5x+4=0$

$(x^{2}+\frac{5}{2}x)=-4/2$

$b=\frac{5}{2}$   $-\frac{5}{2(2)}=(-\frac{5}{4})$   $(-\frac{5}{4})^{2}=\frac{25}{16}$

$(x^{2}+\frac{5}{2}x+\frac{25}{16})=-2+\frac{25}{16}$

$(x+\frac{5}{4})^{2}=\frac{-32+25}{16}$

$(x+\frac{5}{4})^{2}=-\frac{7}{16}$

$\sqrt{(x+\frac{5}{4})^{2}}=\sqrt{-\frac{7}{16}}$

$(x+\frac{5}{4})= \frac{\sqrt{-7}}{4}$

$x= -\frac{5}{4}\pm \frac{\sqrt{-7}}{4}$   or   $x= \frac{-5\pm \sqrt{-7}}{4}$

Another method to solve for the roots of a quadratic equation is using a quadratic formula. Let’s derive the formula:

$ax^{2}+bx+c=0$

$4a^{2}x^{2}+4abx+4ac=0$

$(4a^{2}x^{2}+4abx+b^{2})+4ac-b^{2}=0$

$(2ax+b)^{2}+4ac-b^{2}=0$

$(2ax+b)^{2}=b^{2}-4ac$

$\sqrt{(2ax+b)^{2}}=\sqrt{b^{2}-4ac}$

$2ax+b = \pm \sqrt{b^{2}-4ac}$

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

ILLUSTRATIVE EXAMPLES

Find the value of the following series.

1) $x^{2}+6x+8$                2) $3x^{2}-5x+2$

SOLUTIONS:

1) Identify first the values of a, b, and c.

a = 1          b = 6          c = 8

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-6\pm \sqrt{6^{2}-4(1)(8)}}{2(1)}$

$x=\frac{-6\pm \sqrt{36-32}}{2} = \frac{-6\pm \sqrt{4}}{2} =\frac{-6\pm 2}{2} = -2 \, and \, -4$

2) Identify first the values of a, b, and c.

a = 3               b = -5            c = 2

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$x=\frac{5\pm \sqrt{(-5)^{2}-4(3)(2)}}{2(3)}$

$x=\frac{5\pm \sqrt{25-24}}{6}=\frac{5\pm \sqrt{1}}{6}=\frac{5\pm 1}{6} = 1 \, and \, 2/3$

DISCUSSION

DISCRIMINANT

Discriminant tells the nature of the roots of a quadratic equation. By determining this, it can be helpful in finding the roots. Using discriminant, you will be told ahead of time if the roots will be real or complex. The table below shows the classification of the roots of a quadratic equation.

ILLUSTRATIVE EXAMPLES

Determine the nature of the roots of the following quadratic equations.

1) $x^{2}-8x+1=0$            2) $2x^{2}+5x+4=0$

SOLUTIONS:

1) Remember that is determining the nature of the roots using discriminant, use the formula: $b^{2}-4ac$. Let us first write the value of the coefficients.

a = 1              b = -8               c = 1

$b^{2}-4ac=(-8)^{2}-4(1)(1)=64-4=60$

60>0. Thus the roots of the equation are real and unequal.

2) Let us first write the value of the coefficients.

a = 2              b = 5               c = 4

$b^{2}-4ac=(5)^{2}-4(2)(4)=25-32=-7$

-7<0. Thus the roots of the equation are imaginary and unequal.

DISCUSSION