QUADRATIC EQUATIONS

Examples

Example

Solve the following Quadratic Equation Using Factorization Method:

$\boldsymbol{x^{2} - 7x + 12 = 0}$

Clue:We introduce the brackets from the "sum" and the "product"

Solution

(i)The product = x² Coefficient multiplied by The constant

The $\boldsymbol{x^{2}}$ coefficient is $\boldsymbol{1}$

The Constant is $\boldsymbol{12}$

Product = $\boldsymbol{1 \times 12 = 12}$

(ii)Find the sum

Clue:The sum is the "x" coefficient

In this case, the sum is $\boldsymbol{-7}$

(iii)Find the factors

Ask yourself: Which two numbers gives "-7" when added and "12" when multiplied?

The factors are "-3" and "-4"

This is because $\boldsymbol{(-3) + (-4) = -7}$ and $\boldsymbol{-3 \times -4 = 12}$

(iv)Replace the sum by the factors

$\boldsymbol{x^{2} - 3x - 4x + 12 = 0}$

(v)Factor out the Equation

$\boldsymbol{x(x - 3) - 4(x - 3) = 0}$

After Factorizing, you will notice that 2 brackets are identical. Take only one of them.

Form another bracket from the Equation

The resulting Equation will be :

$\boldsymbol{(x - 3) (x - 4) = 0}$

(vi)Solve each bracket separately.

$\boldsymbol{(x - 3) = 0}$

$\boldsymbol{x = 3}$

you notice that you will find two values of x.

A quadratic equation usually has two solutions (roots). These are denoted as $\boldsymbol{x_{1}}$ and $\boldsymbol{x_{2}}$

Final Answers

Example

Solve the following Quadratic Equation Using Completing Square Method:

$\boldsymbol{x^{2} - 7x + 12 = 0}$

solution

(i)Make sure the equation is in the form:

$\boldsymbol{ax^{2} + bx + c = 0}$

Our Equation is already in that form:

$\boldsymbol{x^{2} - 7x + 12 = 0}$

(ii)Move the constant to the other side

(leave x terms on one side)

$\boldsymbol{x^{2} - 7x = -12}$

(iii)Make sure the coefficient of x² is 1

Divide Through by a (if a ≠ 1)

$\boldsymbol{x^{2} - 7x = -12}$

(iv)Use the formula: "$\boldsymbol{\left(\frac{b}{2}\right)^{2} = c}$" to make another constant for the equation

(Take half of the coefficient of x, then square it)

NOTE : For easier calculation , just Write it as $\boldsymbol{\left(\frac{b}{2}\right)^{2}}$

(v)It becomes: $\boldsymbol{\left(\frac{-7}{2}\right)^{2}}$ which is the same as $\boldsymbol{(-3.5)^{2}}$

(vi)Add that number to both sides of the equation

$\boldsymbol{x^{2} - 7x + (3.5)^{2} = -12 + (-3.5)^{2}}$

(vii)The left side is now a perfect square trinomial, which can be written as: $\boldsymbol{(x - 3.5)^{2}}$

(viii)Simplify the Left side of the equation and evaluate the right part of the equation.

$\boldsymbol{(x - 3.5)^{2} = -12 + 12.25}$

$\boldsymbol{(x - 3.5)^{2} = 0.25}$

(ix)To find "x",we need to find the squareroot of each part.

$\boldsymbol{\sqrt{(x - 3.5)^{2}} = \pm\sqrt{0.25}}$

(x)Solve for x

$\boldsymbol{(x - 3.5) = \pm 0.5}$

$\boldsymbol{x_{1} = + 0.5 + 3.5}$

$\boldsymbol{x_{1} = 4}$

$\boldsymbol{x_{2} = - 0.5 + 3.5}$

$\boldsymbol{x_{2} = 3}$

Final Answers

Example

Solve the following Quadratic Equation Using The Quadratic Formula:

$\boldsymbol{x^{2} - 7x + 12 = 0}$

solution

(i) Make sure the equation is in the form :

$\boldsymbol{ax^{2} + bx + c = 0}$

Our Equation is already in that form:

$\boldsymbol{x^{2} - 7x + 12 = 0}$

(ii) Relate this Equation with : $\boldsymbol{ax^{2} + bx + c = 0}$

After relating, we discover that:

• $\boldsymbol{a = 1}$
• $\boldsymbol{b = -7}$
• $\boldsymbol{c = 12}$

(iii) Substitute the values of "a" , "b" , and "c" to the Quadratic Formula

$\boldsymbol{x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4(1)(12)}}{2(1)}}$

Simplify the solution

$\boldsymbol{x = \frac{7 \pm \sqrt{49 - 48}}{2}}$

$\boldsymbol{x = \frac{7 \pm \sqrt{1}}{2}}$

After simplifying:

$\boldsymbol{x = \frac{7 \pm 1}{2}}$

(iv) Solve for x

$\boldsymbol{x_{1} = \frac{7 + 1}{2}}$

$\boldsymbol{x_{1} = \frac{8}{2}}$

$\boldsymbol{x_{1} = 4}$

$\boldsymbol{x_{2} = \frac{7 - 1}{2}}$

$\boldsymbol{x_{2} = \frac{6}{2}}$

$\boldsymbol{x_{2} = 3}$

Final Answers

Example 1 (Factorization Method)

Solve the following quadratic equation by factorizing: x² - 5x - 14 = 0

Solution

Step 1: Identify the product and the sum.

The product is the coefficient of $\boldsymbol{x^{2}}$ multiplied by the constant:

The coefficient of $\boldsymbol{x^{2} = 1 }$

Constant = −14

Product = $\boldsymbol{1 \times (-14) = -14}$

The sum is the coefficient of $\boldsymbol{x}$:

The coefficient of $\boldsymbol{x = -5 }$

Sum = $\boldsymbol{-5}$

Step 2: Find the factors.

We look for two numbers that multiply to $\boldsymbol{-14}$ (product) and add up to $\boldsymbol{-5}$ (sum).

The factors are $\boldsymbol{-7}$ and $\boldsymbol{2}$ because $\boldsymbol{-7 \times 2 = -14}$ and $\boldsymbol{-7 + 2 = -5}$.

Step 3: Replace the middle term ($\boldsymbol{-5x}$) with the factors and factorize by grouping.

$\boldsymbol{x^{2} - 7x + 2x - 14 = 0}$

Group the first two terms and the last two terms: $\boldsymbol{x(x - 7) + 2(x - 7) = 0}$

Take one of the identical brackets and group the outside terms: $\boldsymbol{(x - 7)(x + 2) = 0}$

Step 4: Solve each bracket separately.

$\boldsymbol{x - 7 = 0 \implies x = 7}$

$\boldsymbol{x + 2 = 0 \implies x = -2}$

Final Answer

$\boldsymbol{x_{1} = 7, x_{2} = -2}$

Example 2 (Factorization with a > 1)

Solve the following quadratic equation using Factorization method : 3x² + 10x + 8 = 0

Solution

Step 1: Identify the product and the sum.

The product is the coefficient of $\boldsymbol{x^{2}}$ multiplied by the constant:

The coefficient of $\boldsymbol{x^{2} = 3 }$

Constant = 8

Product = $\boldsymbol{3 \times 8 = 24}$

The sum is the coefficient of $\boldsymbol{x}$:

The coefficient of $\boldsymbol{x = 10 }$

Sum = $\boldsymbol{10}$

Step 2: Find the factors.

Two numbers that multiply to $\boldsymbol{24}$ and add up to $\boldsymbol{10}$ are $\boldsymbol{6}$ and $\boldsymbol{4}$.

Step 3: Rewrite the equation and factorize by grouping.

$\boldsymbol{3x^{2} + 6x + 4x + 8 = 0}$

$\boldsymbol{3x(x + 2) + 4(x + 2) = 0}$

$\boldsymbol{(x + 2)(3x + 4) = 0}$

Step 4: Solve each bracket separately.

$\boldsymbol{x + 2 = 0 \implies x = -2}$

$\boldsymbol{3x + 4 = 0 \implies 3x = -4 \implies x = \frac{-4}{3}}$

Final Answer

$\boldsymbol{x_{1} = -2, x_{2} = \frac{-4}{3}}$

Example 3 (Completing the Square Method)

Solve the following quadratic equation by completing the square: x² - 6x + 5 = 0

Solution

Step 1: Move the constant term to the right side of the equation.

$\boldsymbol{x^{2} - 6x = -5}$

Step 2: Find the value needed to make a perfect square trinomial using the formula: $\boldsymbol{\left(\frac{b}{2}\right)^{2} = c}$.

Take half of the $\boldsymbol{x}$ coefficient, then square it. For easier calculation, we just write it as: $\boldsymbol{\left(\frac{-6}{2}\right)^{2}}$

Step 3: Add this value to both sides of the equation.

$\boldsymbol{x^{2} - 6x + \left(\frac{-6}{2}\right)^{2} = -5 + \left(\frac{-6}{2}\right)^{2}}$

Since $\boldsymbol{\frac{-6}{2} = -3}$, this simplifies directly to:

$\boldsymbol{x^{2} - 6x + (-3)^{2} = -5 + (-3)^{2}}$

Step 4: Simplify the right part of the equation.

$\boldsymbol{x^{2} - 6x + (-3)^{2} = 4}$

Step 5: Rewrite the left side as a perfect square and take the square root of both sides.

$\boldsymbol{(x - 3)^{2} = 4}$

$\boldsymbol{\sqrt{(x - 3)^{2}} = \pm\sqrt{4}}$

$\boldsymbol{x - 3 = \pm 2}$

Step 6: Solve for both values of $\boldsymbol{x}$.

$\boldsymbol{x_{1} = 2 + 3 = 5}$

$\boldsymbol{x_{2} = -2 + 3 = 1}$

Final Answer

$\boldsymbol{x_{1} = 5, x_{2} = 1}$

Example 4 (Quadratic Formula Method)

Solve the following equation using the quadratic formula: 2x² + 5x - 3 = 0

Solution

Step 1: Identify the constants by relating the expression to $\boldsymbol{ax^{2} + bx + c = 0}$.

We discover that: $\boldsymbol{a = 2, b = 5, c = -3}$

Step 2: Substitute the values into the universal quadratic formula.

The formula is: $\boldsymbol{x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

Substituting values: $\boldsymbol{x = \frac{-5 \pm \sqrt{(5)^{2} - 4(2)(-3)}}{2(2)}}$

Step 3: Show the simplification steps clearly.

$\boldsymbol{x = \frac{-5 \pm \sqrt{25 - (-24)}}{4}}$

$\boldsymbol{x = \frac{-5 \pm \sqrt{25 + 24}}{4}}$

$\boldsymbol{x = \frac{-5 \pm \sqrt{49}}{4}}$

After simplifying: $\boldsymbol{x = \frac{-5 \pm 7}{4}}$

Step 4: Solve for both roots separately.

$\boldsymbol{x_{1} = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}}$

$\boldsymbol{x_{2} = \frac{-5 - 7}{4} = \frac{-12}{4} = -3}$

Final Answer

$\boldsymbol{x_{1} = \frac{1}{2}, x_{2} = -3}$

Example 5 (Word Problem)

The length of a rectangular garden is 3 meters more than its width. If the total area of the garden is 40 m², find the dimensions of the garden.

Solution

Step 1: Formulate a quadratic equation from the given statements.

Let the width be $\boldsymbol{x}$. Therefore, the length will be $\boldsymbol{x + 3}$.

We know that: $\text{Area of a rectangle} = \text{Length} \times \text{Width}$

Therefore :

$\boldsymbol{40 = x(x + 3)}$

Expand the bracket: $\boldsymbol{40 = x^{2} + 3x}$

Rearrange into standard form ($\boldsymbol{ax^{2} + bx + c = 0}$): $\boldsymbol{x^{2} + 3x - 40 = 0}$

Step 2: Solve the equation using factorization method.

$\boldsymbol{x^{2} + 3x - 40 = 0}$

Product = $\boldsymbol{-40}$, Sum = $\boldsymbol{3}$. The factors are $\boldsymbol{8}$ and $\boldsymbol{-5}$.

$\boldsymbol{x^{2} + 8x - 5x - 40 = 0}$

$\boldsymbol{x(x + 8) - 5(x + 8) = 0}$

$\boldsymbol{(x + 8)(x - 5) = 0}$

Step 3: Evaluate the dimensions.

$\boldsymbol{x + 8 = 0 \implies x = -8}$

$\boldsymbol{x_{1} = -8}$

$\boldsymbol{x - 5 = 0 \implies x = 5}$

$\boldsymbol{x_{2} = -5}$

Since measurements of length cannot be negative, we reject $\boldsymbol{-8}$. Therefore, width $\boldsymbol{x = 5\text{ m}}$.

Calculate length: $\boldsymbol{\text{Length} = x + 3 = 5 + 3 = 8\text{ m}}$.

Final Answer

Width = 5 m, Length = 8 m