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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 1.3.2 Quadratic Identities.

Activity 1.3.5.

Work in Groups.
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Form a group of atleast \(4.\) individuals. Define, discuss and work on the following:
  1. Quadratic identities.
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  2. Difference of squares.
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  3. Perfect squares.
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  4. Factorization of quadratic expressions.
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Copy the following expressions and identies, observe and discuss:
  1. \(\displaystyle (a + b)^2 = a^2 + 2ab + b^2\)
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  2. \(\displaystyle (a - b)^2 = a^2 - 2ab + b^2 \)
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  3. \((a - b)(a + b) = a^2 - b^2\) (Difference of squares)
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  • Compare the different approaches groups used to solve similar problems.
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  • Discuss how quadratic identities can make simplification and factoring easier.
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  • Explore how these identities are useful in different contexts (e.g., solving quadratic equations, simplifying expressions in algebra).
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  • How do the identities help us solve quadratic expressions faster?
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  • What happens if we don’t recognize the identity right away—how might that slow us down?
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  • Can you think of any real-world applications where you might use quadratic identities?s
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Suppose we have a equation \(P\) and \(Q.\)
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In mathematics, an equation \(P = Q\) is called an identity if the following conditions are satisfied:
  1. Both sides of the equality relation contain some variables.
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  2. Both the sides give the same value when the variable is substituted with a particular constant.
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Note 1.3.5.

If an equation satisfies the two conditions mentioned above, it is known as an identity. There are many mathematical relations that can be classified as identities, but it’s not necessary to memorize all of them. However, certain key identities in algebra are essential, as they simplify calculations.
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Quadratic identities are special rules or formulas that help us work with quadratic equations.
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Common quadratic identities are:
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Subsubsection 1.3.2.1 Difference of Squares.

Activity 1.3.6.

Work in Groups.
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Define, discuss, and work on the following:
  1. Difference of squares.
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  2. Identifying difference of squares in algebraic expressions.
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  3. Factoring using the difference of squares formula.
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  4. Recognizing patterns when applying the difference of squares.
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Copy the following expressions and identities, observe and discuss:
  1. \(a^2 - b^2 = (a + b)(a - b)\) (Difference of squares)
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Recognize how the difference of squares applies to algebraic expressions.
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  • Compare the different approaches groups used to solve similar problems.
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  • Discuss how the difference of squares formula makes factoring easier.
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  • Explore how this identity is useful in different contexts (e.g., simplifying expressions, solving equations).
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  • How does recognizing the difference of squares help in solving quadratic equations?
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  • What happens if we don’t recognize the difference of squares—how might that affect solving the problem?
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  • Can you think of any real-world applications where the difference of squares might be used?
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The Difference of Squares is an identity that applies when you have two terms that are perfect squares and are being subtracted from each other. The identity states:
\begin{align*} a^2 - b^2 \amp = (a - b)(a + b) \end{align*}
This means that the difference between two squares can be factored into the product of two binomials: one where the terms are subtracted and one where the terms are added.
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In the expression \(a^2 - b^2\text{,}\) \(a\) and \(b\) represent the square roots of the two terms. When you subtract two perfect squares, you can factor the expression as the product of two binomials: \((a - b)\) and \((a + b)\text{.}\)
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Example 1.3.6.

Consider the quadratic, solve:
\begin{align*} \amp x^2 - 16 \end{align*}
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Solution.
The expression \(x^2 - 16\) is in the form of a difference of squares, which follows the identity:
\begin{align*} a^2 - b^2 \amp = (a -b)(a + b) \end{align*}
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In this case, \(a^2 = x^2\) and \(b^2 = 16.\) The square roots of \(x^2\) and \(16\) are \(x\) and \(4,\) respectively. We can rewrite the expression as:
\begin{align*} x^2 - 16 \amp = x^2 - 4^2 \end{align*}
Applying the difference of squares identity we have:
\begin{align*} x^2 - 16 \amp = (x - 4)(x + 4) \end{align*}
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Subsubsection 1.3.2.2 Perfect Squares.

Activity 1.3.7.

Work in Groups.
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Define, discuss, and work on the following:
  1. Perfect square identities.
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  2. Expanding perfect squares.
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  3. Recognizing perfect square trinomials.
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  4. Factoring perfect square trinomials.
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Copy the following expressions and identities, observe and discuss:
  1. \(\displaystyle (a + b)^2 = a^2 + 2ab + b^2\)
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  2. \(\displaystyle (a - b)^2 = a^2 - 2ab + b^2\)
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Recognize the middle term is twice the product of the two terms.
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  • Compare the different approaches groups used to solve similar problems.
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  • Discuss how perfect square identities can make simplification and factoring easier.
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  • Explore how these identities are useful in different contexts (e.g., solving quadratic equations, simplifying expressions in algebra).
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  • How do perfect square identities help us solve quadratic expressions faster?
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  • What happens if we don’t recognize the identity right away—how might that slow us down?
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  • Can you think of any real-world applications where you might use perfect square identities?
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A Perfect Square is a special kind of trinomial that can be factored into the square of a binomial. There are two forms of perfect square identities:
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\begin{align*} a^2 + 2ab +b^2 \amp = (a + b)^2 \end{align*}
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\begin{equation*} and \end{equation*}
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\begin{align*} a^2 - 2ab + b^2 \amp = (a - b)^2 \end{align*}
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In these identities, the expression on the left is a perfect square, which means it can be written as the square of a binomial (a two-term expression). The first identity is used when the middle term is positive, and the second identity is used when the middle term is negative.
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Example 1.3.7. (Using the first identity).

Consider the quadratic below:
\begin{equation*} x^2 + 6x + 9 \end{equation*}
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Solution.
The expression \(x^2 + 6x + 9\) is perfefect square. A perfect square follows the pattern:
\begin{align*} (a + b)^2 \amp = a^2 + 2ab + b^2 \end{align*}
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Here, \(a^2 = x^2, 2ab = 6x\) and \(b^2 = 9.\) The square roots of \(x^2\) and \(9\) are \(x\) and \(3,\) respectively.
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In this case, the expression \(x^2 + 6x + 9\) matches the form \(a^2 + 2ab + b^2,\) where:
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The middle term \(2ab\) is \(6x,\) which fits the formula for a perfect square with a negative middle term. Applying the identity:
\begin{align*} x^2 + 6x + 9 \amp = (x + 3)^2 \end{align*}
So, \(x^2 + 6x + 9\) factors into \((x + 3)^2.\)
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Example 1.3.8. (Using the second identity).

Using the quadratic below:
\begin{equation*} x^2 - 10x + 25 \end{equation*}
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Solution.
The expression \(x^2 - 10x + 25\) matches the form \(a^2 - 2ab + b^2\) where:
  • \(\displaystyle a = x,\)
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  • \(\displaystyle b = 5.\)
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  • The middle term \(2ab = -10x,\) and the constant term \(b^2 = 25.\)
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So, this is a perfect square, and we can factor it as:
\begin{align*} x^2 - 10x + 25 \amp = (x - 5)^2 \end{align*}
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Subsubsection 1.3.2.3 Factoring Quadratic.

Activity 1.3.8.

Work in Groups.
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Define, discuss, and work on the following:
  1. Factorization of quadratic expressions.
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  2. Identifying common factors in expressions.
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  3. Factorizing using the method of splitting the middle term.
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  4. Recognizing the difference between factoring by grouping and simple factoring.
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Copy the following expressions, observe and discuss:
  1. Factorize: \(x² + 5x + 6\)
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  2. Factorize: \(x² - 7x + 12\)
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  3. Factorize: \(3x² - 15x\)
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  4. Factorize by grouping: \(x² + 4x + 3x + 12\)
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  • Compare the different approaches groups used to factor similar expressions.
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  • Discuss how factoring helps in solving quadratic equations.
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  • Explore how recognizing common factors and patterns makes factoring easier.
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  • How does factoring help us simplify algebraic expressions faster?
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  • What challenges do you face when factoring complex expressions?
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  • Can you think of any real-world scenarios where factoring is useful (e.g., optimizing areas, engineering problems)?
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Factoring quadratic expressions involves expressing a quadratic equation in the form of two binomials. A general quadratic equation looks like:
\begin{equation*} (ax + m)(bx + n) \end{equation*}
To factor a quadratic, we find two numbers \(m\) and \(n\) such that:
  1. \(m \times n = ac\) (the product of \(a\) and \(c\)),
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  2. \(m + n = b\) (the coefficient of \(x\))
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Once we find \(m\) and \(n\text{,}\) we break the middle term \(bx\) into two terms using \(m\) and \(n\text{,}\) then factor by grouping.
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Example 1.3.9.

Cosider:
\begin{equation*} x^2 + 5x + 6 \end{equation*}
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Solution.
In this example \(a = 1, b = 5\) and \(c = 6.\) We need two numbers that when multiplied it gives \(ac = 1 \times 6 = 6\) and when the the two values are added up they give \(b = 5.\) These two numbers are \(2\) and \(3.\)
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\begin{equation*} 2 \times 3 = 6 \end{equation*}
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\begin{equation*} \text{and} \end{equation*}
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\begin{equation*} 2 + 3 = 5 \end{equation*}
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Now, rewriting the middle term \(5x\) as \(2x + 3x,\) and then factor by grouping
\begin{align*} x^2 + 2x + 3x + 6 \amp \end{align*}
Grouping the terms:
\begin{align*} (x^2 + 2x) + (3x + 6) \amp \end{align*}
Factoring each groups or movng common factor of each and create brackets:
\begin{align*} x(x + 2) + 3(x + 2) \amp \end{align*}
Now, factor out the common binomials:
\begin{align*} (x + 2)(x + 3) \amp \end{align*}
Finally, \(x^2 + 5x + 6\) factors to \((x + 2)(x + 3).\)
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Exercises 1.3.2.4 Exercises

1.

Use the quadratic identities to write down the expansions of each of the following expressions:
  1. \(\displaystyle (4x + 5)^2\)
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  2. \(\displaystyle \left( \frac{1}{x} + \frac{1}{y} \right) \left( \frac{1}{x} - \frac{1}{y} \right)\)
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  3. \(\displaystyle (8 - x)^2\)
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  4. \(\displaystyle (x - 7)^2\)
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  5. \(\displaystyle \left( x + \frac{1}{2} \right)^2\)
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  6. \(\displaystyle \left( \frac{1}{4} - \frac{3}{4}b \right)^2\)
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  7. \(\displaystyle (x + 2)^2\)
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  8. \(\displaystyle (x + 5)^2\)
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  9. \(\displaystyle (x + 2)(x + 3)\)
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