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Subsection 1.3.5 Solutions Of Quadratic Equations By Factorisations.

Activity 1.3.11.

Work in Groups.
Find the roots of the equation using factorization.
  1. \(x² - 5x + 6 = 0\text{.}\)
  2. \(x² + 7x + 10 = 0\text{.}\)
After solving the equations, discuss the following:
  • What steps did you follow in factorizing the quadratic equation?
  • How do the roots relate to the factors of the quadratic equation?
  • Did any group use a different method to factorize the equation? Compare approaches.
Finally, each group will share their solved quadratic equations and the methods they used infront of the classroom.
In our previous sections we have discussed the form of a quadratic expression, identities and quadratic equations; how to form quadratic equations and differentiate between an expression and equation
In this section we are going to learn about solving the factored equation, we are going to continue from the point where we form factored equation
Numbers that satisfy an equation (its solutions) are called the roots of the equation.
Once you have factored the quadratic into the form;
\begin{align*} (x+p)(x+q) \amp = 0 \end{align*}
Set each factor equal to zero and solve for \(x\text{:}\)
\begin{align*} (x + p) \amp = 0 \end{align*}
\begin{align*} \text{or} \amp \end{align*}
\begin{align*} (x + q) \amp = 0 \end{align*}
Solving these will give the two solutions for \(x\text{.}\)

Example 1.3.17.

Solve
\begin{align} x^2 + 5x + 6 \amp = 0 \tag{1.3.1} \end{align}
Solution.
Look for two numbers that if multiplied by \(6\) and when added up gives \(5\)
These numbers are \(2\) and \(3.\) So we can write the equation as:
\begin{align*} x^2 + 2x + 3x + 6 \amp = 0 \end{align*}
Finding a common factor we have:
\begin{align*} x(x + 2) + 3(x + 2) \amp = 0 \end{align*}
Grouping we have th factors of the equation as:
\begin{align*} (x +2) (x + 3) \amp = 0 \end{align*}
Setting each factor equal to zero (0).
\begin{equation*} x + 2 = 0 \end{equation*}
\begin{equation*} \text{or} \end{equation*}
\begin{equation*} x + 3 = 0 \end{equation*}
Solving for \(x.\)
\begin{equation*} x = -2 \end{equation*}
\begin{equation*} \text{or} \end{equation*}
\begin{equation*} x = -3 \end{equation*}
Thus, the solutions to the quadratic equation \(x^+5x+6=0\) are;
\begin{equation*} x = (-2) \, \text {and} \, x = (-3). \end{equation*}

Example 1.3.18.

Solve,
\begin{align} 6x^2 + 13x + 6 \amp = 0 \tag{1.3.2} \end{align}
Solution.
We need 2 numbers to form factors, which when;
These numbers are \(9\) and \(4.\) So rewrite the middle part.
\begin{align*} 6x^2 + 9x + 4x + 6 \amp = 0 \end{align*}
Grouping the factors:
\begin{align*} (6x^2 + 9x) + (4x + 6) \amp = 0 \end{align*}
Factor each group:
\begin{align*} 3x(2x + 3) + 2(2x + 3) \amp = 0 \end{align*}
\(\,\)
\begin{align*} (2x+3)(3x+2) \amp = 0 \end{align*}
Set each factor eaqual to zero:
\begin{equation*} 2x + 3 = 0 \end{equation*}
\begin{equation*} \text{or} \end{equation*}
\begin{equation*} 3x + 2 = 0 \end{equation*}
Solving for \(x.\)
\begin{equation*} x = -\frac{3}{2} \end{equation*}
\begin{equation*} \text{or} \end{equation*}
\begin{equation*} x = -\frac{2}{3} \end{equation*}
Thus, the solutions to \(6x^2 +13x+6=0\) are:
\begin{equation*} x = -\frac{3}{2} \, \text{and} \, x = -\frac{2}{3} \end{equation*}

Exercises Exercises

1.

Solve the quadratic equation by factorisation.
  1. \(\displaystyle x^2 + 7x + 10 = 0\)
  2. \(\displaystyle x^2 - 5x + 6 = 0\)
  3. \(\displaystyle x^2 + 3x - 4 = 0\)
  4. \(\displaystyle 2x^2 + 9x + 7 = 0\)
  5. \(\displaystyle x^2 - 6x + 8 = 0\)

2.

  1. A car’s speed is represented by a quadratic equation:
    \begin{align*} 4x^2 - 16x + 15 \amp = 0 \text{.} \end{align*}
    Find the possible values of x representing time.
  2. The sum of a number and its square is \(42\text{.}\) Form a quadratic equation and solve it to find the number.
  3. Solve the quadratic equation:
    \begin{equation*} 3x^2 - 14x + 8 = 0 \end{equation*}
  4. A garden’s area is 56 square meters, and its length is 4 meters more than its width. Form and solve a quadratic equation to find the dimensions of the garden.