2.2b Solving Quadratic Equations – Completing the Square

Posted on April 18, 2020 by user

2.4.2 Solving Quadratic Equations – Completing the Square

(A) The Perfect Square

1. The expression x² + 2x + 1 can be written in the form (x + 1)², it is called a “perfect square”.

2. If the algebraic expression on the left hand side of the quadratic equation is a perfect, the roots can be easily obtained by finding the square roots.

Example:
Solve each of the following quadratic equation

(a) (x + 1)² = 25

(b) x² − 8x + 16 = 49

Solution:

(a)

(x + 1)² = 25

(x + 1)² = ±√25

x = −1 ± 5

x = 5 or x = −6

(b)

x² − 8x + 16 = 49

(x − 4)² = 49

(x − 4) = ±√49

x = 4 ± 7

x = 11 or x = −3

(B) Solving Quadratic Equation by Completing the Square

1. To solve quadratic equation, we make the left hand side of the equation a perfect square.

2. To make any quadratic expression x² + px into a perfect square, we add the term (p/2)² to the expression and this will make

3. The following shows the steps to solve the equation by using completing the square method for quadratic equation ax²+ bx = – c.

(a) Rewrite the equation in the form ax² + bx = – c.

(b) If the coefficient a ≠ 1, reduce it to 1 (by dividing).

(c) Add (p/2)² to both sides of the equation.

(d) Write the expression on the left hand side as a perfect square.

(e) Solve the equation.

Solving Quadratic Equations

Posted on April 18, 2020 by user

Solving Quadratic Equations

To solve a quadratic equation means to find all the roots of the quadratic equations.

Example:
Find the roots of the quadratic equations
a.

x^{2} = 9

b.

2 x^{2} - 98 = 0

Answer:
(a)

\begin{array}{l} x^{2} = 9 \\ x = \pm \sqrt{9} \\ x = \pm 3 \end{array}

(b)

\begin{array}{l} 2 x^{2} - 98 = 0 \\ 2 x^{2} = 98 \\ x^{2} = \frac{98}{2} = 49 \\ x = \pm \sqrt{49} = \pm 7 \end{array}

A quadratic equation may be solve by using one of the following method

Factorisation
Completing the square
Using quadratic formula

Roots of Quadratic Equation – Example

Posted on April 18, 2020 by user

Remember : Roots of a quadratic equation are the values of variables/unknowns that satisfy the equation.

Example
(a) Given x = 3 is the root of the quadratic equation

x^{2} + 2 x + p = 0

, find the value of p.
(b) The roots of the quadratic equation

3 x^{2} + h x + k = 0

are -2 and 4. Find the value of h and k.

General Form of Quadratic Equation

Posted on April 18, 2020 by user

General Form of Quadratic Equation

General form of a quadratic equation is

a x^{2} + b x + c = 0

where a, b, and c are constants and a≠0.

*Note that the highest power of an unknown of a quadratic equation is 2.

Example (Find the values of a, b and c)
Rewrite the following into the general form of a quadratic equation. Find the values of a, b, and c.
(a)

{(3 x - 5)}^{2} = 0

(b) (x - 8)(x + 8) = 0

Solution

Quadratic Equations (Introduction)

Posted on April 18, 2020 by user

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of the second degree.
A quadratic equation has only one variable
The highest power of the variable is 2.

Example of Quadratic Equation
The followings are some examples of quadratic equations

$2 x^{2} + 3 x + 4 = 0$
$t^{2} = \frac{2}{5}$
$y (6 y - 3) = 5$
$a - 3 y + 2 y^{2} = 2$ where a is a constant.

Example of Non Quadratic Equation

$2 x + 1 = 0$ , (Reason: The highest power of x ≠ 2.)
$2 x^{3} + 1 = x$ , (Reason: The highest power of x ≠ 2.)
$t^{2} + \frac{5}{t} = 3$ , (Reason: The present of the term $\frac{5}{t}$ .)

SPM Practice (Long Question)

Posted on April 18, 2020 by user

Question 5:
In diagram below, the function g maps set P to set Q and the function h maps set Q to set R.

Find
(a) in terms of x, the function
(i) which maps set Q to set P,
(ii) h(x).
(b) the value of x such that gh(x) = 8x + 1.

Solution:
(a)(i)

\begin{array}{l} g (x) = 3 x + 2 \\ Let g^{- 1} (x) = y \\ g (y) = x \\ 3 y + 2 = x \\ y = \frac{x - 2}{3} \\ g^{- 1} (x) = \frac{x - 2}{3} \end{array}

(a)(ii)

\begin{array}{l} h g (x) = 12 x + 5 \\ h (3 x + 2) = 12 x + 5 \to g (x) = 3 x + 2 \\ Let u = 3 x + 2 \\ x = \frac{u - 2}{3} \\ h (u) = 12 (\frac{u - 2}{3}) + 5 \\ = 4 u - 8 + 5 \\ = 4 u - 3 \\ h (x) = 4 x - 3 \end{array}

(b)

\begin{array}{l} g h (x) = g (4 x - 3) \\ = 3 (4 x - 3) + 2 \\ = 12 x - 9 + 2 \\ = 12 x - 7 \\ 12 x - 7 = 8 x + 1 \\ 4 x = 8 \\ x = 2 \end{array}

Inverse Function Example 7

Posted on April 18, 2020 by user

Example 7 (Comparison Method)
Given that

f : x \to \frac{2 h}{x - 3 k}

, x≠3k , where h and k are constants and

f^{- 1} : x \to \frac{14 + 24 x}{x}

, x≠0, find the value of h and of k.

Solution:

Inverse Function Example 6 (Comparison Method)

Posted on April 18, 2020 by user

Example 6 (Comparison Method)
If

f : x \mapsto \frac{m x - n}{x - 2}, x \neq 2

and

f^{- 1} : x \mapsto \frac{5 - 2 x}{2 - x}, x \neq 2.

. Find the value of m and of n,

Inverse Function Example 5

Posted on April 18, 2020 by user

Example 5

If $g : x \mapsto \frac{m - x}{x - 3}, x \neq 3$ and $g^{- 1} (5) = 14$ . Find the value of m.

Inverse Function Example 3

Posted on April 18, 2020 by user

Example 3
Find the inverse function of

f (x) = \frac{3 x + 2}{5 x + 3}