Monday, July 20, 2026

The Ultimate Guide to Percentage Increase and Percentage Decrease for Digital SAT Math, GCSE, IGCSE, and ACT

The Ultimate Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Mathematics (Part 2)

Understanding Percentage Increase and Percentage Decrease from First Principles

In the previous chapter, you learned that a percentage represents a quantity out of every one hundred equal parts. You also learned how to convert between fractions, decimals and percentages and how to calculate a percentage of a number.

In this chapter, we will build on those ideas to understand percentage increase and percentage decrease. Instead of memorising formulas, you will learn why these calculations work. Once you understand the reasoning behind the mathematics, percentage problems become much easier, whether they appear in Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge Mathematics, Edexcel Mathematics, AQA Mathematics, or any other secondary school mathematics course.

Although examination questions are written in different styles, the mathematical ideas remain exactly the same.

Why Do We Use Percentages to Compare Change?

Imagine two libraries.

Library A receives 40 new books.

Library B also receives 40 new books.

At first glance, both libraries appear to have grown by exactly the same amount.

However, suppose Library A originally had 200 books, while Library B originally had 2,000 books.

The increase is identical.

The effect is not.

Adding forty books to a collection of two hundred is a much greater change than adding forty books to a collection of two thousand.

This is why mathematicians do not compare only the increase.

Instead, they compare the increase with the original quantity.

Once this comparison has been made, the result is converted into a percentage.

Using percentages allows us to compare changes fairly, even when the original quantities are completely different.

Building the Formula Yourself

Suppose a quantity increases.

There are three important numbers.

• The original value.

• The new value.

• The increase.

The increase is found first.

Increase = New Value − Original Value

Now compare the increase with the original quantity.

Increase ÷ Original Value

This comparison gives a decimal.

Since percentages mean "out of every one hundred," multiply the decimal by 100.

The complete calculation becomes

Percentage Increase = (Increase ÷ Original Value) × 100%

Notice that this formula has been developed logically instead of being memorised.

Example 1

A reading club had 160 members at the beginning of the year.

By the end of the year, it had 200 members.

Find the percentage increase.

Step 1

Find the increase.

Increase

= 200 − 160

= 40

The club gained forty new members.

Step 2

Compare the increase with the original number.

40 ÷ 160

= 0.25

This means the increase is one quarter of the original membership.

Step 3

Convert the decimal into a percentage.

0.25 × 100

= 25%

Answer

The membership increased by 25%.

Why Do We Divide by the Original Value?

Suppose another reading club also gained forty members.

This club increased from 800 members to 840 members.

Again,

Increase

= 40

Now compare it with the original size.

40 ÷ 800

= 0.05

0.05 × 100

= 5%

Although both clubs gained forty members, the percentage increase is very different.

First club

25%

Second club

The original value determines how significant the increase really is.

Example 2

A wildlife park recorded 480 visitors on Saturday.

On Sunday, 600 visitors entered the park.

Find the percentage increase.

Step 1

Find the increase.

600 − 480

= 120

Step 2

Divide by the original number.

120 ÷ 480

= 0.25

Step 3

Convert to a percentage.

0.25 × 100

= 25%

Answer

The number of visitors increased by 25%.

Understanding Percentage Decrease

A percentage decrease follows exactly the same reasoning.

The only difference is that the quantity becomes smaller instead of larger.

Again, we compare the amount of change with the original quantity.

Building the Formula

First calculate the decrease.

Decrease

= Original Value − New Value

Next compare this decrease with the original value.

Decrease ÷ Original Value

Finally convert the decimal into a percentage.

This gives

Percentage Decrease = (Decrease ÷ Original Value) × 100%

Notice that the denominator has not changed.

The original quantity is always used because that is where the change began.

Example 3

A nature reserve contained 950 trees.

After a severe storm,

874 trees remained.

Find the percentage decrease.

Step 1

Calculate the decrease.

950 − 874

= 76

Step 2

Compare with the original number.

76 ÷ 950

= 0.08

Step 3

Convert to a percentage.

0.08 × 100

= 8%

Answer

The number of trees decreased by 8%.

Example 4

A water tank originally contained 1,500 litres of water.

After irrigation,

1,200 litres remained.

Find the percentage decrease.

Step 1

Decrease

= 1,500 − 1,200

= 300

Step 2

Compare with the original quantity.

300 ÷ 1,500

= 0.2

Step 3

Convert to a percentage.

0.2 × 100

= 20%

Answer

The amount of water decreased by 20%.

Finding the New Value After a Percentage Increase

Sometimes the percentage increase is given instead of the new value.

Instead of finding the percentage change, your task is to calculate the new quantity.

There are two reliable methods.

The first method develops the answer step by step.

The second method uses a multiplier.

Both methods produce exactly the same result.

Example 5

Increase 640 by 15%.

Method 1

Find 15% of 640.

15 × 640

= 9,600

Now divide by 100.

9,600 ÷ 100

= 96

The increase is 96.

Now add this increase to the original value.

640 + 96

= 736

Therefore,

the new value is 736.

Method 2

A 15% increase means the final quantity becomes

100% + 15%

= 115%

Convert 115% into a decimal.

115%

= 1.15

Now multiply.

640 × 1.15

= 736

Both methods produce exactly the same answer.

Why Does the Multiplier Work?

Many students memorise multipliers without understanding them.

Suppose a quantity increases by 12%.

The original quantity already represents

100%.

Adding another 12% gives

112%.

Since

112%

= 112 ÷ 100

= 1.12

Multiplying by

1.12

automatically includes both the original quantity and the increase.

Understanding this idea makes multipliers much easier to remember.

SAT Strategy

Many examination questions never use the words percentage increase or percentage decrease. Instead, they describe situations involving attendance, rainfall, production, business sales, scientific experiments, test scores, or population changes. Before performing any calculation, identify the original quantity, the new quantity, and whether the change represents an increase or a decrease. This simple habit helps prevent many common errors and is especially useful in Digital SAT Math, ACT Math, GCSE Mathematics, and IGCSE Mathematics.

Practice Questions

A museum welcomed 720 visitors on Monday and 900 visitors on Tuesday. Find the percentage increase.
A reservoir contained 4,500 cubic metres of water. After a dry season, it contained 3,960 cubic metres. Find the percentage decrease.
Increase 840 by 18%.
Increase 360 by 12.5%.
A factory produced 2,400 bicycles last year and 2,760 this year. Find the percentage increase.
A theatre sold 640 tickets on Friday and 560 on Saturday. Find the percentage decrease.

Answers

25%
12%
991.2
405
15%
12.5%

Chapter Summary

Percentage increase and percentage decrease measure how much a quantity changes relative to its original value. By comparing the amount of change with the starting quantity, percentages provide a fair way of comparing situations involving different sizes. This principle is widely used in mathematics, science, economics, business, finance, statistics, and data analysis, making it an essential skill for success in the Digital SAT, PSAT, ACT, GCSE, IGCSE, and other secondary mathematics examinations.

In the next chapter, you will explore successive percentage changes, discounts, profit and loss, sales tax, compound percentage change, and advanced percentage word problems that combine several mathematical concepts into a single question.

Finding the New Value After a Percentage Decrease Using a Multiplier

The multiplier method works just as well when a quantity decreases.

Instead of adding the percentage to 100%, subtract the percentage from 100%.

The remaining percentage represents the portion of the original quantity that is left.

Building the Multiplier

Suppose a quantity decreases by 18%.

The original quantity represents

100%

Subtract the decrease.

100% − 18%

= 82%

Now convert 82% into a decimal.

82%

= 82 ÷ 100

= 0.82

Therefore, 0.82 is the multiplier.

Instead of calculating the decrease separately and subtracting it afterwards, you can simply multiply the original quantity by 0.82.

Example 6

A tablet originally costs ₹24,000.

The store offers a 15% discount.

Find the sale price using the multiplier method.

Step 1

Calculate the multiplier.

100% − 15%

= 85%

Convert 85% into a decimal.

85%

= 85 ÷ 100

= 0.85

Step 2

Multiply the original price by the multiplier.

24,000 × 0.85

= 20,400

Therefore,

the sale price is ₹20,400.

Example 7

A library contained 3,200 books.

After removing damaged books, the collection decreased by 12.5%.

How many books remained?

Step 1

Find the multiplier.

100% − 12.5%

= 87.5%

Convert to a decimal.

87.5%

= 0.875

Step 2

Multiply.

3,200 × 0.875

= 2,800

Therefore,

2,800 books remained in the library.

Example 8

A warehouse stored 960 boxes.

After shipping 35% of them, how many boxes remained?

Step 1

Find the multiplier.

100% − 35%

= 65%

Convert to a decimal.

65%

= 0.65

Step 2

Multiply.

960 × 0.65

= 624

Therefore,

624 boxes remained in the warehouse.

Friday, July 17, 2026

Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Algebra

Master Percentages with Step by Step Explanations, Worked Examples

Introduction

Percentages are one of the most useful mathematical concepts you will ever learn. Whether you are calculating a discount while shopping, comparing examination scores, interpreting statistical reports, reading graphs, analysing scientific data, or solving algebra problems, percentages appear almost everywhere.

In mathematics examinations, percentage questions often look simple, but they frequently test several concepts at the same time. A single problem may combine percentages with fractions, decimals, ratios, equations, graphs, probability, data analysis, or financial mathematics. Learning to recognise these connections is an important step towards becoming a confident problem solver.

A solid understanding of percentages is valuable for students preparing for the Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge IGCSE Mathematics, Edexcel GCSE Mathematics, AQA GCSE Mathematics, OCR GCSE Mathematics, Scottish National 5 Mathematics, and many other secondary school mathematics courses around the world. Although examination styles may differ, the mathematical principles remain exactly the same.

This guide has been written from first principles. Every method is explained carefully, every algebraic step is shown, and every worked example follows a logical sequence so that you understand why each step works instead of simply memorising a formula.

After studying this chapter, you will be able to

• Understand the meaning of a percentage.

• Convert between percentages, fractions and decimals.

• Find the percentage of any quantity.

• Determine what percentage one number is of another.

• Solve percentage increase and percentage decrease problems.

• Apply percentage concepts to algebra and word problems.

• Develop the mathematical reasoning required for college entrance examinations and secondary school mathematics.

What Does the Word Percentage Mean?

The word percentage comes from the Latin phrase meaning per hundred.

Therefore,

1% means 1 out of every 100 equal parts.

Similarly,

10% means 10 parts out of 100.

25% means 25 parts out of 100.

75% means 75 parts out of 100.

100% means the entire quantity.

Understanding this simple idea makes every percentage calculation much easier.

Writing Percentages as Fractions

Every percentage can be written as a fraction whose denominator is 100.

Examples

25%

= 25/100

= 1/4

50%

= 50/100

= 1/2

75%

= 75/100

= 3/4

80%

= 80/100

= 4/5

125%

= 125/100

= 5/4

Notice that percentages greater than 100% are perfectly possible. They simply represent quantities larger than the original amount.

Writing Percentages as Decimals

Many Digital SAT, ACT, GCSE and IGCSE questions require changing percentages into decimals.

The rule is simple.

Divide the percentage by 100.

Examples

45%

= 45 ÷ 100

= 0.45

= 8 ÷ 100

= 0.08

150%

= 150 ÷ 100

= 1.5

0.5%

= 0.5 ÷ 100

= 0.005

Moving the decimal point two places to the left produces exactly the same result.

Converting Decimals into Percentages

To change a decimal into a percentage,

multiply by 100.

Examples

0.6

= 0.6 × 100

= 60%

0.08

= 0.08 × 100

= 8%

1.25

= 1.25 × 100

= 125%

Always remember to write the percentage symbol after multiplying by 100.

Converting Fractions into Percentages

There are two common methods.

Method 1

Convert the fraction into a decimal first.

Example

3/5

Divide.

3 ÷ 5

= 0.6

Multiply by 100.

0.6 × 100

= 60%

Method 2

Multiply the fraction directly by 100.

Example

3/5 × 100

= 300/5

= 60%

Both methods produce the same answer.

Choose whichever method you find easier.

Finding the Percentage of a Number

One of the most common examination questions asks you to calculate a certain percentage of a quantity.

The general rule is

Percentage of a number = Percentage × Number ÷ 100

Example 1

Find 25% of 80.

Step 1

Write the formula.

Percentage of a number

= Percentage × Number ÷ 100

Step 2

Substitute the values.

25 × 80 ÷ 100

Step 3

Multiply.

25 × 80

= 2000

Step 4

Divide by 100.

2000 ÷ 100

= 20

Therefore,

25% of 80 is 20.

Example 2

Find 18% of 250.

Step 1

Write the formula.

Percentage × Number ÷ 100

Step 2

Substitute.

18 × 250 ÷ 100

Step 3

Multiply.

18 × 250

= 4500

Step 4

Divide.

4500 ÷ 100

= 45

Therefore,

18% of 250 equals 45.

Example 3

Find 12.5% of 96.

Step 1

Write the formula.

Percentage × Number ÷ 100

Step 2

Substitute.

12.5 × 96 ÷ 100

Step 3

Multiply.

12.5 × 96

= 1200

Step 4

Divide.

1200 ÷ 100

= 12

Therefore,

12.5% of 96 is 12.

Using Fractions Instead of Percentages

Sometimes converting the percentage into a fraction makes the calculation much faster.

Example

Find 50% of 240.

50%

= 1/2

Half of 240

= 120

No multiplication is necessary.

Find 25% of 64.

25%

= 1/4

One quarter of 64

= 16

Find 75% of 80.

75%

= 3/4

First find one quarter.

80 ÷ 4

= 20

Now multiply by 3.

20 × 3

= 60

This approach is often quicker during timed examinations.

What Percentage Is One Number of Another?

Another common examination question asks

"What percentage is one quantity of another?"

The formula is

Percentage

= (Part ÷ Whole) × 100

Example 4

A class contains 40 students.

Twenty-eight students passed an examination.

What percentage passed?

Step 1

Identify the part.

Step 2

Identify the whole.

Step 3

Use the formula.

(28 ÷ 40) × 100

Step 4

Divide.

28 ÷ 40

= 0.7

Step 5

Multiply.

0.7 × 100

= 70%

Therefore,

70% of the students passed the examination.

Example 5

A football team won 18 matches out of 24.

What percentage of matches did they win?

Step 1

Write the formula.

(Part ÷ Whole) × 100

Step 2

Substitute.

(18 ÷ 24) × 100

Step 3

Simplify.

18 ÷ 24

= 0.75

Step 4

Multiply.

0.75 × 100

= 75%

Therefore,

The team won 75% of its matches.

SAT Strategy

Many Digital SAT, ACT, GCSE and IGCSE questions disguise percentage problems inside word problems, graphs, tables or algebraic expressions. Before beginning any calculation, identify whether the question is asking you to find a percentage of a quantity, what percentage one quantity is of another, or how much a quantity changes by a given percentage. Recognising the type of problem before performing any arithmetic often saves valuable time during an examination.

Practice Questions

Find 35% of 240.
Find 12% of 350.
Find 62.5% of 160.
Express 7/20 as a percentage.
Express 0.84 as a percentage.
What percentage is 45 out of 60?
What percentage is 18 out of 48?
Find 5% of 640.
Find 125% of 48.
A school has 600 students. If 456 students attend on a particular day, what percentage attended?

Answers

84
42
100
35%
84%
75%
37.5%
32
60
76%

Solve the following initial value differential equation (x − 1) dy/dx = 2xy, when y(2) = 1

Solve the following initial value differential equation

(x − 1) dy/dx = 2xy, when y(2) = 1.

This is a variable separable differential equation.

Separate variables

(x − 1) dy/dx = 2xy

dy/y = [2x / (x − 1)] dx

Integrate both sides

∫ dy/y = ∫ [2x / (x − 1)] dx

2x/(x−1) = 2 + 2/(x−1) using long division or manipulation of the numerator

∫ dy/y =∫ [2 + 2/(x − 1)] dx

ln|y| = 2x + 2ln|x − 1| + C

Apply initial condition y(2) = 1

When x = 2, y = 1

ln|1| = 2(2) + 2ln|2 − 1| + C

0 = 4 + 2ln(1) + C

0 = 4 + 0 + C

⇒ C = −4

ln|y| = 2x + 2ln|x − 1| − 4

ln|y| - 2ln|x − 1| = 2x − 4

using property of loagarithms

ln|y| - ln|x − 1|² =2x − 4

ln [|y| / |x − 1|² ] =2x − 4

y = (x − 1)² e^(2x − 4)

see this video for more explanation

cbse 12th applied mathematics variable separable differential equation previous year question papers 2025 2026

Thursday, July 16, 2026

The Coordinates of the Centre of a Circle Are (x − 7, 2x): Find the Value of x if the Circle Passes Through (−9, 11) and Has Radius 5√2 | Step-by-Step Solution

The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of ‘x’, if the circle passes through the point (−9, 11) and has radius 5√2 units.

For a circle,

distance between centre and any point on circle = radius

using square of distance formula:

(x₂ − x₁)² + (y₂ − y₁)² = r²

Given

Centre = (x − 7, 2x)

Point on circle = (−9, 11)

Radius r = 5√2

r² = (5√2)² = 25 × 2 = 50

(-9 - (x - 7))² + (11 - 2x)² = 50

(-9 - x + 7)² + (11 - 2x)² = 50

(-x - 2)² + (11 - 2x)² = 50

(x + 2)² + (11 - 2x)² = 50

Expand using identities

(x² + 4x + 4) + (121 - 44x + 4x²) = 50

5x² - 40x + 125 = 50

5x² - 40x + 125 - 50 =0

5x² - 40x + 75 = 0

Divide by 5

x² - 8x + 15 = 0

Factorise:

x² - 5x - 3x + 15 = 0

x(x - 5) - 3(x - 5) = 0

(x - 5)(x - 3) = 0

x = 5 or x = 3

for more explanation watch the video

cbse 10th maths coordinate geometry distance formula previous year question paper 2025 2026

Friday, July 10, 2026

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

A + B + C together fill the tank in 8 hours

So, rate of (A + B + C) = [1/8] tank per hour

Work done by A + B + C in 2 hours = 2 × 1/8 =[ 1/4 ]tank

Remaining work = 1 − 1/4 = [3/4 ]tank

Remaining [3/4] tank is filled by A + C in 9 hours

So, rate of (A + C) = 3/4 × 1/9 = 1/12 tank per hour

Rate of B = Rate of (A + B + C) − Rate of (A + C)

= 1/8 − 1/12

= 3/24 − 2/24 =[ 1/24 [tank per hour

Time taken by B alone = 24 hours

see this video for more explanation

pipes problem, cbse 12th applied maths old board exam question paper 2025 2026

Solving Quadratic Equations by Completing the Square The Digital SAT Math Guide to Quadratic Equations (Part 2)

The Digital SAT Math Guide to Quadratic Equations (Part 2)

Solving Quadratic Equations by Completing the Square

In the previous chapter, you learned how to solve quadratic equations by factoring. Factoring is often the quickest method, but many quadratic equations on the Digital SAT cannot be factored easily. Some have large coefficients, some produce fractional values, and others have no integer factors at all.

For these equations, completing the square provides a systematic method that always works. Unlike factoring, you do not have to guess factor pairs or recognize patterns. Instead, you follow the same sequence of algebraic steps every time.

Before learning the procedure, remember one important rule.

The coefficient of x² should be 1 before you begin completing the square.

A quadratic equation whose coefficient of x² is 1 is called a monic quadratic equation.

If the equation is not monic, divide every term on both sides of the equation by the coefficient of x². This makes the remaining steps much easier and reduces mistakes.

Universal Method for Completing the Square

Always follow these steps.

Write the equation in standard form.
If the coefficient of x² is not 1, divide every term on both sides by that coefficient.
Move the constant term to the opposite side.
Take half of the coefficient of x.
Square that number.
Add the squared value to both sides.
Rewrite the left side as the square of a binomial.
Take the square root of both sides.
Remember both the positive and negative square roots.
Solve the resulting linear equations.
Check every solution in the original equation.

Example 1

Solve

2x² + 12x + 4 = 0

Step 1

The coefficient of x² is 2.

Divide every term on both sides by 2.

2x² ÷ 2 + 12x ÷ 2 + 4 ÷ 2 = 0 ÷ 2

Simplify.

x² + 6x + 2 = 0

The equation is now monic.

Step 2

Subtract 2 from both sides.

x² + 6x + 2 − 2 = 0 − 2

Simplify.

x² + 6x = −2

Step 3

Take half of 6.

6 ÷ 2 = 3

Square it.

3² = 9

Step 4

Add 9 to both sides.

x² + 6x + 9 = −2 + 9

Simplify.

x² + 6x + 9 = 7

Step 5

Rewrite the left side.

(x + 3)² = 7

Step 6

Take square roots.

√((x + 3)²) = ±√7

Simplify.

x + 3 = ±√7

Step 7

Subtract 3 from both sides.

Positive solution:

x = −3 + √7

Negative solution:

x = −3 − √7

Example 2

Now solve a question that produces fractions immediately after making the quadratic monic.

4x² + 10x − 3 = 0

Step 1

The coefficient of x² is 4.

Divide every term by 4.

4x² ÷ 4 + 10x ÷ 4 − 3 ÷ 4 = 0 ÷ 4

Simplify.

x² + ⁵⁄₂x − ³⁄₄ = 0

Notice that fractions are perfectly acceptable. Do not convert them to decimals because exact fractions make later calculations more accurate.

Step 2

Move the constant term.

Add ³⁄₄ to both sides.

x² + ⁵⁄₂x = ³⁄₄

Step 3

Take half of the coefficient of x.

The coefficient is ⁵⁄₂.

Half of ⁵⁄₂ is

⁵⁄₂ ÷ 2 = ⁵⁄₄

Now square the result.

(⁵⁄₄)² = ²⁵⁄₁₆

Step 4

Add ²⁵⁄₁₆ to both sides.

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁄₄ + ²⁵⁄₁₆

Convert ³⁄₄ to sixteenths.

³⁄₄ = ¹²⁄₁₆

Now add.

¹²⁄₁₆ + ²⁵⁄₁₆ = ³⁷⁄₁₆

The equation becomes

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁷⁄₁₆

Step 5

Rewrite the left side.

(x + ⁵⁄₄)² = ³⁷⁄₁₆

Step 6

Take square roots.

√((x + ⁵⁄₄)²) = ±√(³⁷⁄₁₆)

Simplify.

x + ⁵⁄₄ = ±√37⁄4

Step 7

Subtract ⁵⁄₄ from both sides.

x = −⁵⁄₄ ± √37⁄4

These are the exact solutions.

Notice that completing the square works just as well with fractions as it does with whole numbers. On the Digital SAT, leaving answers in exact fractional or radical form is often the correct approach unless the question specifically asks for a decimal approximation.

Thursday, July 9, 2026

The Digital SAT Math Guide to Quadratic Equations (Part 1)

Master Quadratic Equations for the Digital SAT with Step-by-Step Explanations

Quadratic equations are one of the most important algebra topics on the Digital SAT. They appear in many forms, from straightforward equation-solving questions to graph interpretation, mathematical modeling, and real-world word problems. A strong understanding of quadratics also makes it much easier to learn functions, parabolas, coordinate geometry, and polynomial expressions.

Unlike linear equations, which produce straight lines when graphed, quadratic equations create curved graphs called parabolas. Learning how these equations behave will help you answer a wide variety of SAT Math questions quickly and accurately.

This guide is written for students who want to build a solid understanding of quadratics from the ground up. Every solution is explained one step at a time, with no skipped steps or unexplained shortcuts. By the time you finish this chapter, you'll understand what quadratic equations are, how to recognize them, and how to solve many of them by factoring.

Learning Goals

In this chapter, you will learn how to:

Recognize a quadratic equation.
Understand why quadratic equations are different from linear equations.
Identify the standard form of a quadratic equation.
Understand quadratic expressions and quadratic functions.
Solve simple quadratic equations by factoring.
Apply the Zero Product Property.
Check your answers correctly.
Avoid common mistakes made on the Digital SAT.

These concepts form the foundation for more advanced methods such as completing the square and using the quadratic formula, which will be covered in later chapters.

What Is a Quadratic Equation?

A quadratic equation is an equation in which the highest exponent of the variable is 2.

Examples include:

x² = 49

x² + 5x + 6 = 0

2x² − 7x + 3 = 0

4x² = 100

Notice that each equation contains x².

That squared variable is what makes the equation quadratic.

Compare these two equations.

Linear equation:

2x + 7 = 13

Highest exponent = 1

Quadratic equation:

x² + 2x − 15 = 0

Highest exponent = 2

The difference may seem small, but it changes how the equation behaves. A linear equation usually has one solution, while a quadratic equation can have two solutions, one solution, or no real solutions.

The Standard Form of a Quadratic Equation

Most quadratic equations on the SAT are written in standard form:

ax² + bx + c = 0

Each letter has a meaning.

a is the coefficient of x².

b is the coefficient of x.

c is the constant term.

For example,

3x² + 8x − 11 = 0

Here,

a = 3

b = 8

c = −11

Learning to identify these three values is important because later methods, especially the quadratic formula, use them directly.

Understanding the Parts of a Quadratic Equation

Consider

2x² + 9x − 18 = 0

This equation has three terms.

First term:

2x²

This is called the quadratic term because it contains x².

Second term:

This is called the linear term because it contains x.

Third term:

−18

This is the constant term because it contains no variable.

Recognizing these parts helps you identify which solving method to use.

What Does It Mean to Solve a Quadratic Equation?

Solving a quadratic equation means finding every value of the variable that makes the equation true.

For example,

x² = 25

Which numbers produce 25 when squared?

5² = 25

(−5)² = 25

Therefore,

x = 5

and

x = −5

Unlike linear equations, quadratic equations often have more than one correct answer.

Why Are There Two Answers?

Many students are surprised to discover that one equation can have two solutions.

The reason is simple.

Squaring removes the negative sign.

Positive example:

5 × 5 = 25

Negative example:

−5 × −5 = 25

Both calculations produce the same answer.

Whenever you solve an equation involving x², always ask yourself whether both a positive and a negative solution are possible.

Introduction to Factoring

Factoring x² + bx + c type when a = 1

Factoring is one of the fastest methods for solving many quadratic equations on the Digital SAT.

Factoring means rewriting an expression as the product of two smaller expressions.

Example:

x² + 5x + 6

can be written as

(x + 2)(x + 3)

When multiplied together,

(x + 2)(x + 3)

= x² + 3x + 2x + 6

= x² + 5x + 6

The original expression and its factored form are mathematically identical.

The Zero Product Property

Factoring works because of an important algebra rule.

A × B = 0

then

A = 0

B = 0

or both.

This rule is called the Zero Product Property.

Example:

(x + 4)(x − 7) = 0

Either

x + 4 = 0

x − 7 = 0

Solve each equation separately.

First equation:

x + 4 = 0

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Second equation:

x − 7 = 0

Add 7 to both sides.

x − 7 + 7 = 0 + 7

Simplify.

x = 7

Therefore,

the two solutions are

x = −4

and

x = 7

Example 1

Solve

x² + 7x + 12 = 0

Step 1

Write the equation.

x² + 7x + 12 = 0

Step 2

Find two numbers whose product is 12 and whose sum is 7.

Possible factor pairs of 12 are

1 and 12

2 and 6

3 and 4

Only

3 and 4

add to 7.

Step 3

Write the factors.

(x + 3)(x + 4) = 0

Step 4

Apply the Zero Product Property.

Either

x + 3 = 0

x + 4 = 0

Step 5

Solve the first equation.

Subtract 3 from both sides.

x + 3 − 3 = 0 − 3

Simplify.

x = −3

Step 6

Solve the second equation.

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Final Answer

x = −3

x = −4

Example 2

Solve

x² − 9x + 20 = 0

Step 1

Find two numbers whose product is 20.

1 and 20

2 and 10

4 and 5

Step 2

Which pair adds to −9?

Since the product is positive and the sum is negative,

both numbers must be negative.

−4 and −5

Step 3

Write the factors.

(x − 4)(x − 5) = 0

Step 4

Set each factor equal to zero.

x − 4 = 0

x − 5 = 0

Step 5

Solve.

Add 4 to both sides.

x = 4

Add 5 to both sides.

x = 5

Check

4² − 9(4) + 20

16 − 36 + 20

Correct.

Now check 5.

25 − 45 + 20

Correct.

Both answers satisfy the equation.

Example 3

Solve

x² + x − 12 = 0

Step 1

Find two numbers whose product is −12.

Possible pairs include

1 and −12

2 and −6

3 and −4

Step 2

Find the pair whose sum equals 1.

4 and −3

Step 3

Write the factors.

(x + 4)(x − 3) = 0

Step 4

Set each factor equal to zero.

x + 4 = 0

x − 3 = 0

Step 5

Solve.

Subtract 4 from both sides.

x = −4

Add 3 to both sides.

x = 3

Final Answer

x = −4

x = 3

A Quick Factoring Strategy

Whenever you see

x² + bx + c

ask yourself two questions.

Question 1

Which two numbers multiply to give c?

Question 2

Do those same numbers add to give b?

If the answer is yes,

you have found the correct factors.

With practice, this process becomes much faster.

Forgetting to Check

Substitute every solution back into the original equation.

If the equation balances,

your solution is correct.

Practice Questions

Solve by factoring.

x² + 5x + 6 = 0
x² − 8x + 15 = 0
x² + 9x + 20 = 0
x² − 7x + 10 = 0
x² + 2x − 15 = 0
x² − x − 12 = 0
x² + 11x + 24 = 0
x² − 10x + 24 = 0

Answers

x = −2, −3
x = 3, 5
x = −4, −5
x = 2, 5
x = 3, −5
x = 4, −3
x = −3, −8
x = 4, 6

A quadratic equation is an equation whose highest exponent is two. Before attempting to solve it, identify whether it is already in standard form and determine the values of a, b, and c. When the equation can be factored, rewriting it as the product of two binomials often provides the quickest solution. The Zero Product Property then allows each factor to be solved separately, producing all possible solutions. As you continue practicing, you'll begin to recognize common factor patterns quickly, an essential skill for success on the Digital SAT Math section.