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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 1.3.4 Formation of Quadratic Equations by Factorisations.

Activity 1.3.10.

Work in Groups.
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  1. Start by forming a quadratic equation using given roots.
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  2. Write your equation in factorized form.
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  3. Swap your equation with another group, solve it and find the roots.
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After completing the task, discuss these questions with the class:
  • How did you find the quadratic equation from the roots?
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  • What did you learn about factorization from this exercise?
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  • Did you notice any patterns when forming quadratic equations?
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Finally, each group will share their quadratic equation and the method used with the class.
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Any equation of the form \(ax^2 + bx + c = 0\) where \(a, b \, \text{and} \, c\) are constants and \(a \neq 0\) is known as a quadratic equation. In this section, solutions of quadratic equations using the factor method is discussed.
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To solve a quadratic equation by factorisation, we aim to rewrite the quadratic equation in the form of:
\begin{align*} (x+p)(x+q) \amp = 0 \end{align*}
where \(p\) and \(q\) are numbers that, when multiplied, give the product \(c\) (the constant term), and when added, give the sum \(b\) (the coefficient of the middle term).
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To solve a quadratic eaquation, first ensure the equations is in the form;
\begin{equation*} ax^2 + bx + c = 0. \end{equation*}
if not in the form rearrange, so that all terms are on one side of the equation and the equation equals zero.
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Factor the quadratic expression:
  • Look for two numbers that when multiplied give \(a \times c\) (the coefficient of \(x^2\))
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  • And, when added give \(b\) (the coefficient of the middle term).
    • If the quadratic has a leading coefficient \(a = 1\text{,}\) you only need to find two numbers that multiply to \(c\) and add to \(b\text{.}\)
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    • If \(a \neq 1\text{,}\) find two numbers that multiply to \(a \times c\) and add to \(b\text{.}\)
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Rewrite the middle term using these two numbers and factor by grouping, which you will have a quadratic of the form:
\begin{equation*} (x+p)(x+q) = 0 \end{equation*}
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Example 1.3.15.

Solve
\begin{equation*} x^2 + 6x + 8 = 0 \end{equation*}
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Solution.
Factoring the left hand side (L.H.S) of the equation.
\begin{equation*} x^2 + 6x + 8 = 0 \end{equation*}
The factors are \(2 \, \text {and} 4\text{.}\)
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\begin{align*} \amp x^2 + 2x + 4x + 8 = 0 \end{align*}
Now create common factors
\begin{align*} \amp x(x + 2) + 4(x + 2) = 0 \end{align*}
Grouping them together we have:
\begin{align*} \amp (x + 2) + (x + 4) = 0 \end{align*}
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Example 1.3.16.

Form a quadratic expression using the factoring method.
\begin{equation*} 6x^2 + 12x \end{equation*}
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Solution.
To form a quadratic expression using the factoring method, first set the expression equal to \(0\) to form a quadratic equation:
\begin{align*} 6x^2 + 12x \amp = 0 \end{align*}
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Remove the common factor out and form a bracket:
\begin{align*} 6x(x + 2) \amp = 0 \end{align*}
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Thus, the factored form of expression \(6x^2 + 12x\) is;
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\begin{align*} 6x(x + 2) \amp = 0 \end{align*}
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Exercises Exercises

1.

Form a quadratic expression using the following expressions.
    1. \(\displaystyle (x + 4)(x + 5)\)
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    2. \(\displaystyle (x + 3)^2\)
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    3. \(\displaystyle 4x^2 + 8x\)
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    4. \(\displaystyle 3x^2 + 6x\)
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    5. \(\displaystyle (p - q)(p - q)\)
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    6. \(\displaystyle (ax + b)(2ax - 3b)\)
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    1. \(\displaystyle (x - 8)(x + 8)\)
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    2. \(\displaystyle (4 - 2x)(\frac{1}{2}x + 3)\)
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    3. \(\displaystyle (5dx + 3d)(2dx - 4d)\)
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    4. \(\displaystyle (\frac{1}{2} + x)^2\)
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    5. \(\displaystyle (\frac{1}{8} + \frac{1}{x})^2\)
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