Quadratic Equations and Methods of Solution for SS 1

Quadratic Equations

Performance Objectives

By the end of the lesson, students should be able to:

1. factorize quadratic expressions;
2. form perfect square quadratic expression with missing constants;
3. derive the quadratic formula and use it to solve a quadratic equation;
4. find the roots of quadratic equations;
5. form quadratic equation with given roots;
6. draw quadratic equation graph;
7. obtain roots from the graph;
8. apply quadratic equation knowledge to real life situation.

Introduction to Quadratic Equations

In mathematics, an equation is a statement that two expressions are equal. A quadratic equation is a specific type of equation that involves a variable raised to the power of two, but no higher power.

The standard form of a quadratic equation is given by:

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

where $x$ is the variable, and $a$, $b$, and $c$ are constants. It is important that $a \neq 0$, because if $a$ were zero, the $x^2$ term would disappear, and the equation would become a linear equation ($bx + c = 0$). The constants $b$ and $c$ can be zero.

Examples of quadratic equations include:

1. $x^2 – 5x + 6 = 0$ (Here, $a=1, b=-5, c=6$)
2. $2y^2 + 7y = 0$ (Here, $a=2, b=7, c=0$)
3. $3p^2 – 12 = 0$ (Here, $a=3, b=0, c=-12$)
4. $m^2 = 9$ (This can be rewritten as $m^2 – 9 = 0$, so $a=1, b=0, c=-9$)

Factorization of Quadratic Expressions

A quadratic expression is an algebraic expression of the form $ax^2 + bx + c$, where $a \neq 0$. Factorization is the process of rewriting a quadratic expression as a product of two linear expressions (binomials).

To factorize a quadratic expression $ax^2 + bx + c$, the following steps are generally followed:

1. Multiply the coefficient of $x^2$ (which is $a$) by the constant term (which is $c$). Let this product be $P = ac$.
2. Find two numbers, let’s call them $m$ and $n$, such that their product is $P$ ($m \times n = ac$) and their sum is the coefficient of $x$ (which is $b$) ($m + n = b$).
3. Rewrite the middle term $bx$ as the sum of $mx$ and $nx$. The expression becomes $ax^2 + mx + nx + c$.
4. Group the terms into two pairs: $(ax^2 + mx) + (nx + c)$.
5. Factor out the common factor from each pair. The expressions in the brackets should be identical.
6. Factor out the common binomial expression.

Worked Examples

Example 1: Factorize $x^2 + 5x + 6$.

1. Here, $a=1, b=5, c=6$. The product $ac = 1 \times 6 = 6$.
2. Find two numbers whose product is 6 and sum is 5. These numbers are 2 and 3 ($2 \times 3 = 6$ and $2 + 3 = 5$).
3. Rewrite the expression: $x^2 + 2x + 3x + 6$.
4. Group the terms: $(x^2 + 2x) + (3x + 6)$.
5. Factor out common factors: $x(x + 2) + 3(x + 2)$.
6. Factor out the common binomial: $(x + 2)(x + 3)$.

Therefore, $x^2 + 5x + 6 = (x + 2)(x + 3)$.

Example 2: Factorize $2x^2 – 7x + 3$.

1. Here, $a=2, b=-7, c=3$. The product $ac = 2 \times 3 = 6$.
2. Find two numbers whose product is 6 and sum is -7. These numbers are -1 and -6 ($-1 \times -6 = 6$ and $-1 + (-6) = -7$).
3. Rewrite the expression: $2x^2 – x – 6x + 3$.
4. Group the terms: $(2x^2 – x) – (6x – 3)$. (Note the sign change when factoring out a negative).
5. Factor out common factors: $x(2x – 1) – 3(2x – 1)$.
6. Factor out the common binomial: $(2x – 1)(x – 3)$.

Therefore, $2x^2 – 7x + 3 = (2x – 1)(x – 3)$.

Solving Quadratic Equations by Factorization

The principle for solving quadratic equations by factorization relies on the zero product property: if the product of two or more factors is zero, then at least one of the factors must be zero.

This means if $ab = 0$, then $a = 0$ or $b = 0$ (or both). This property is fundamental to solving quadratic equations by factorization.

To solve a quadratic equation $ax^2 + bx + c = 0$ by factorization:

1. Ensure the equation is in standard form, with one side equal to zero.
2. Factorize the quadratic expression on the other side into two linear factors.
3. Set each linear factor equal to zero.
4. Solve each resulting linear equation for the variable. The solutions are called the roots of the quadratic equation.

Worked Examples

Example 1: Solve $x^2 – 5x + 6 = 0$ by factorization.

1. The equation is already in standard form.
2. Factorize the quadratic expression $x^2 – 5x + 6$.
   • Product $ac = 1 \times 6 = 6$.
   • Numbers whose product is 6 and sum is -5 are -2 and -3.
   • $x^2 – 2x – 3x + 6 = 0$
   • $x(x – 2) – 3(x – 2) = 0$
   • $(x – 2)(x – 3) = 0$
3. Set each factor to zero:
   • $x – 2 = 0$ or $x – 3 = 0$
4. Solve for $x$:
   • $x = 2$ or $x = 3$

The roots of the equation are $x = 2$ and $x = 3$.

Example 2: Solve $3x^2 + 10x – 8 = 0$ by factorization.

1. The equation is in standard form.
2. Factorize $3x^2 + 10x – 8$.
   • Product $ac = 3 \times (-8) = -24$.
   • Numbers whose product is -24 and sum is 10 are 12 and -2.
   • $3x^2 + 12x – 2x – 8 = 0$
   • $3x(x + 4) – 2(x + 4) = 0$
   • $(3x – 2)(x + 4) = 0$
3. Set each factor to zero:
   • $3x – 2 = 0$ or $x + 4 = 0$
4. Solve for $x$:
   • $3x = 2 \implies x = \frac{2}{3}$
   • $x = -4$

The roots of the equation are $x = \frac{2}{3}$ and $x = -4$.

Perfect Square Quadratic Expressions

A perfect square quadratic expression is a quadratic expression that can be factored into two identical linear factors. It is of the form $(px + q)^2$ or $(px – q)^2$.

When expanded, these take the form:

1. $(px + q)^2 = (px)^2 + 2(px)(q) + q^2 = p^2x^2 + 2pqx + q^2$
2. $(px – q)^2 = (px)^2 – 2(px)(q) + q^2 = p^2x^2 – 2pqx + q^2$

Notice that the constant term ($q^2$) is always positive, and it is the square of half the coefficient of $x$ (when the coefficient of $x^2$ is 1).

To form a perfect square quadratic expression with a missing constant, consider the expression $x^2 + bx + k$. For this to be a perfect square, the constant term $k$ must be equal to the square of half the coefficient of $x$.

$$k = \left(\frac{b}{2}\right)^2$$

Worked Examples

Example 1: Find the constant $k$ that makes $x^2 + 8x + k$ a perfect square.

1. The coefficient of $x$ is $b=8$.
2. Half of $b$ is $\frac{8}{2} = 4$.
3. Square this value: $k = 4^2 = 16$.

So, $x^2 + 8x + 16$ is a perfect square, which can be written as $(x + 4)^2$.

Example 2: Find the constant $k$ that makes $x^2 – 10x + k$ a perfect square.

1. The coefficient of $x$ is $b=-10$.
2. Half of $b$ is $\frac{-10}{2} = -5$.
3. Square this value: $k = (-5)^2 = 25$.

So, $x^2 – 10x + 25$ is a perfect square, which can be written as $(x – 5)^2$.

Deriving and Using the Quadratic Formula

Not all quadratic equations can be easily solved by factorization. The quadratic formula provides a general method to find the roots of any quadratic equation in the standard form $ax^2 + bx + c = 0$.

Derivation of the Quadratic Formula

The quadratic formula can be derived using the method of completing the square.
Consider the standard quadratic equation:

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

Step 1: Divide by $a$ (assuming $a \neq 0$).

$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

Step 2: Move the constant term to the right side.

$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$

Step 3: Complete the square on the left side.
To do this, add $\left(\frac{1}{2} \times \text{coefficient of } x\right)^2$ to both sides.
The coefficient of $x$ is $\frac{b}{a}$. Half of this is $\frac{b}{2a}$. Squaring it gives $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$.

$$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$$

Step 4: Factor the left side as a perfect square and simplify the right side.

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} – \frac{c}{a}$$
$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 – 4ac}{4a^2}$$

Step 5: Take the square root of both sides. Remember to include both positive and negative roots.

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

Step 6: Isolate $x$.

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

Combining the terms on the right side gives the quadratic formula:

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

Using the Quadratic Formula

To use the quadratic formula, identify the values of $a$, $b$, and $c$ from the quadratic equation $ax^2 + bx + c = 0$, and substitute them into the formula.

Worked Examples

Example 1: Solve $2x^2 + 5x – 3 = 0$ using the quadratic formula.

1. Identify $a, b, c$:
   • $a = 2$
   • $b = 5$
   • $c = -3$
2. Substitute these values into the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
   $$x = \frac{-5 \pm \sqrt{5^2 – 4(2)(-3)}}{2(2)}$$
   $$x = \frac{-5 \pm \sqrt{25 – (-24)}}{4}$$
   $$x = \frac{-5 \pm \sqrt{25 + 24}}{4}$$
   $$x = \frac{-5 \pm \sqrt{49}}{4}$$
   $$x = \frac{-5 \pm 7}{4}$$
3. Calculate the two possible roots:
   • $x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$
   • $x_2 = \frac{-5 – 7}{4} = \frac{-12}{4} = -3$

The roots are $x = \frac{1}{2}$ and $x = -3$.

Example 2: Solve $x^2 + 4x + 1 = 0$ using the quadratic formula.

1. Identify $a, b, c$:
   • $a = 1$
   • $b = 4$
   • $c = 1$
2. Substitute these values into the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
   $$x = \frac{-4 \pm \sqrt{4^2 – 4(1)(1)}}{2(1)}$$
   $$x = \frac{-4 \pm \sqrt{16 – 4}}{2}$$
   $$x = \frac{-4 \pm \sqrt{12}}{2}$$
   $$x = \frac{-4 \pm \sqrt{4 \times 3}}{2}$$
   $$x = \frac{-4 \pm 2\sqrt{3}}{2}$$
3. Simplify by dividing by 2:
   $$x = -2 \pm \sqrt{3}$$

The roots are $x = -2 + \sqrt{3}$ and $x = -2 – \sqrt{3}$.

Finding the Roots of Quadratic Equations

The roots of a quadratic equation are the values of the variable that satisfy the equation, meaning they make the equation true. These roots are also known as the solutions of the equation.

As demonstrated in the previous sections, the roots of a quadratic equation can be found using two primary methods at this level:

1. Factorization: This method is effective when the quadratic expression can be easily factored into two linear binomials. The zero product property is then applied to find the roots.
2. Quadratic Formula: This is a universal method that can solve any quadratic equation, regardless of whether it is easily factorable or not. It provides the roots directly by substituting the coefficients $a, b, c$.

Both methods yield the same roots for a given quadratic equation.

Forming Quadratic Equations with Given Roots

If the roots of a quadratic equation are known, the equation itself can be formed. Let the roots be $\alpha$ (alpha) and $\beta$ (beta).

If $x = \alpha$ and $x = \beta$ are the roots, then:

1. $x – \alpha = 0$
2. $x – \beta = 0$

Using the zero product property in reverse, the quadratic equation can be written as the product of these factors set to zero:

$$(x – \alpha)(x – \beta) = 0$$

Expanding this product gives:

$$x^2 – \beta x – \alpha x + \alpha \beta = 0$$
$$x^2 – (\alpha + \beta)x + \alpha \beta = 0$$

This means that a quadratic equation can be formed as $x^2 – (\text{sum of roots})x + (\text{product of roots}) = 0$.

Worked Examples

Example 1: Form a quadratic equation whose roots are 2 and 3.

1. Let $\alpha = 2$ and $\beta = 3$.
2. Using the factor form: $(x – 2)(x – 3) = 0$.
3. Expand the expression:
   $x(x – 3) – 2(x – 3) = 0$
   $x^2 – 3x – 2x + 6 = 0$
   $x^2 – 5x + 6 = 0$

Alternatively, using the sum and product of roots:

1. Sum of roots: $\alpha + \beta = 2 + 3 = 5$.
2. Product of roots: $\alpha \beta = 2 \times 3 = 6$.
3. Substitute into $x^2 – (\alpha + \beta)x + \alpha \beta = 0$:
   $x^2 – (5)x + (6) = 0$
   $x^2 – 5x + 6 = 0$

The quadratic equation is $x^2 – 5x + 6 = 0$.

Example 2: Form a quadratic equation whose roots are $\frac{1}{2}$ and $-4$.

1. Let $\alpha = \frac{1}{2}$ and $\beta = -4$.
2. Using the factor form: $\left(x – \frac{1}{2}\right)(x – (-4)) = 0$.
3. $\left(x – \frac{1}{2}\right)(x + 4) = 0$.
4. Expand the expression:
   $x(x + 4) – \frac{1}{2}(x + 4) = 0$
   $x^2 + 4x – \frac{1}{2}x – 2 = 0$
   $x^2 + \frac{8x – x}{2} – 2 = 0$
   $x^2 + \frac{7}{2}x – 2 = 0$
5. To remove the fraction, multiply the entire equation by 2:
   $2(x^2) + 2\left(\frac{7}{2}x\right) – 2(2) = 2(0)$
   $2x^2 + 7x – 4 = 0$

Alternatively, using the sum and product of roots:

1. Sum of roots: $\alpha + \beta = \frac{1}{2} + (-4) = \frac{1}{2} – \frac{8}{2} = -\frac{7}{2}$.
2. Product of roots: $\alpha \beta = \frac{1}{2} \times (-4) = -2$.
3. Substitute into $x^2 – (\alpha + \beta)x + \alpha \beta = 0$:
   $x^2 – \left(-\frac{7}{2}\right)x + (-2) = 0$
   $x^2 + \frac{7}{2}x – 2 = 0$
4. Multiply by 2 to clear the fraction:
   $2x^2 + 7x – 4 = 0$

The quadratic equation is $2x^2 + 7x – 4 = 0$.

Drawing Quadratic Graphs

The graph of a quadratic equation $y = ax^2 + bx + c$ is a curve called a parabola. The shape of the parabola depends on the sign of the coefficient $a$.

1. If $a > 0$, the parabola opens upwards (U-shape).
2. If $a < 0$, the parabola opens downwards (inverted U-shape).

To draw a quadratic graph, follow these steps:

1. Create a table of values: Choose a range of $x$-values (e.g., from -3 to 3 or -5 to 5, depending on the equation) and calculate the corresponding $y$-values using the equation $y = ax^2 + bx + c$.
2. Choose appropriate scales for the axes: Decide on suitable scales for the $x$-axis and $y$-axis based on the range of values in your table.
3. Plot the points: Plot each $(x, y)$ pair from your table on the graph paper.
4. Draw a smooth curve: Connect the plotted points with a smooth, continuous curve. Do not use a ruler to connect the points with straight lines.

Worked Example

Example: Draw the graph of $y = x^2 – 2x – 3$ for $x$ values from -2 to 4.

Step 1: Create a table of values.

The points to plot are: $(-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), (4, 5)$.

Step 2: Choose scales.
For the $x$-axis, 2 cm to 1 unit.
For the $y$-axis, 2 cm to 1 unit.

Step 3 & 4: Plot points and draw a smooth curve.

Obtaining Roots from the Graph

The roots of a quadratic equation $ax^2 + bx + c = 0$ are the $x$-values where the graph of $y = ax^2 + bx + c$ intersects the $x$-axis. At these points, the $y$-value is zero.

To obtain the roots from a quadratic graph:

1. Draw the graph of the quadratic equation $y = ax^2 + bx + c$.
2. Identify the points where the curve crosses or touches the $x$-axis.
3. Read the $x$-coordinates of these intersection points. These $x$-coordinates are the roots of the equation.

Example using the previous graph

From the graph of $y = x^2 – 2x – 3$ (shown above), observe where the curve intersects the $x$-axis.

The graph crosses the $x$-axis at $x = -1$ and $x = 3$.

Therefore, the roots of the equation $x^2 – 2x – 3 = 0$ are $x = -1$ and $x = 3$.

Application of Quadratic Equations to Real-Life Situations

Quadratic equations are used to model various real-life situations, especially those involving areas, projectile motion, and optimisation problems. They help in finding unknown quantities that satisfy certain conditions.

To apply quadratic equations to real-life situations:

1. Understand the problem: Read the problem carefully and identify the knowns and unknowns.
2. Define variables: Assign a variable (e.g., $x$) to the unknown quantity that needs to be found.
3. Formulate the equation: Translate the problem’s conditions into a mathematical equation. This will often result in a quadratic equation.
4. Solve the quadratic equation: Use factorization or the quadratic formula to find the roots of the equation.
5. Interpret the solution: Check if the roots make sense in the context of the real-life problem. Sometimes, one or both roots might not be physically possible (e.g., negative length, negative time). Select the appropriate root(s) and state the answer clearly with units.

Worked Examples

Example 1: The length of a rectangular plot of land is 5 metres more than its width. If the area of the plot is 84 square metres, find the dimensions of the plot.

1. Understand the problem: We need to find the length and width of a rectangle given a relationship between them and its area.
2. Define variables:
   Let the width of the plot be $w$ metres.
   Since the length is 5 metres more than its width, the length will be $(w + 5)$ metres.
3. Formulate the equation:
   The area of a rectangle is length $\times$ width.
   Area $= (w + 5) \times w = 84$
   $w^2 + 5w = 84$
   $w^2 + 5w – 84 = 0$ (This is a quadratic equation)
4. Solve the quadratic equation: We can use factorization.
   • Product $ac = 1 \times (-84) = -84$.
   • Numbers whose product is -84 and sum is 5 are 12 and -7.
   • $w^2 + 12w – 7w – 84 = 0$
   • $w(w + 12) – 7(w + 12) = 0$
   • $(w – 7)(w + 12) = 0$
   • So, $w – 7 = 0$ or $w + 12 = 0$
   • $w = 7$ or $w = -12$
5. Interpret the solution:
   Since width cannot be negative, $w = -12$ is not a valid solution for this problem.
   Therefore, the width $w = 7$ metres.
   The length $= w + 5 = 7 + 5 = 12$ metres.

The dimensions of the plot are 7 metres by 12 metres.

Example 2: A stone is thrown vertically upwards from the ground. Its height $h$ (in metres) after $t$ seconds is given by the formula $h = 20t – 5t^2$. After how many seconds will the stone hit the ground?

1. Understand the problem: We need to find the time ($t$) when the stone hits the ground.
2. Define variables:
   $h$ is the height, $t$ is the time.
3. Formulate the equation:
   When the stone hits the ground, its height $h$ is 0.
   So, we set $h = 0$ in the formula:
   $0 = 20t – 5t^2$
   Rearranging into standard form:
   $5t^2 – 20t = 0$ (This is a quadratic equation)
4. Solve the quadratic equation: We can factor out $5t$.
   $5t(t – 4) = 0$
   So, $5t = 0$ or $t – 4 = 0$
   $t = 0$ or $t = 4$
5. Interpret the solution:
   $t = 0$ represents the time when the stone was thrown from the ground.
   $t = 4$ represents the time when the stone hits the ground after its flight.

The stone will hit the ground after 4 seconds.