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EPISTLE 5: QUADRATIC EQUATIONS
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Things You Would Learn

In this epistle, you will learn:

• What a quadratic equation really is, in plain language.

• Where quadratic equations came from, and the people behind them.

• Why the quadratic formula works. The proof, not just the formula to memorize.

• How to solve by factorization, by completing the square, and by the formula.

• How to read the discriminant and know how many roots to expect before you solve.

• How to turn a word problem into a quadratic equation, and check your answer.

• How to work faster, with fewer mistakes, and solve problems you have never seen before.

Start from the basics
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A quadratic equation is an equation where the highest power of the unknown is 2.

Take x² − 5x + 6 = 0. The x² term is what makes it quadratic. If the x² term were gone, you would be back to a linear equation like the ones in Epistle 4, with one answer. The x² term is what gives a quadratic equation up to two answers.

Every quadratic equation can be written in standard form:

ax² + bx + c = 0, where a is not zero.

Here a is the number in front of x², b is the number in front of x, and c is the number on its own. In x² − 5x + 6 = 0, a = 1, b = −5 and c = 6. In 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2.

The answers are called roots. A root is a value of x that makes the equation true. x² − 5x + 6 = 0 has two roots, x = 2 and x = 3, because 2² − 5 × 2 + 6 = 0 and 3² − 5 × 3 + 6 = 0.

Why two roots and not one? Because a square has two square roots. The number 9 is both 3² and (−3)², so whenever you take a square root while solving, two values appear. That is the whole reason a quadratic equation can have two answers. It can also have one repeated answer, or no real answer at all, and the discriminant in this epistle tells you which one to expect before you start.

A brief history and the people behind it
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Quadratic equations are about four thousand years old.

The Babylonians, around 2000 BCE, were already solving problems that we would write as quadratic equations today. They had no symbols and no formula. They worked with lengths and areas, and they solved by completing the square in pictures, cutting and moving rectangles to make a perfect square.

The first clear rules for solving quadratics came from India. Brahmagupta, writing around 628 CE, gave rules for solving quadratic equations and saw that a square root has two values, so an equation can have two answers. He was one of the first to take negative numbers seriously in his arithmetic, though like most mathematicians of his time he did not accept a negative answer as the final solution to a real problem.

The book that gave the subject its method came from Baghdad. Muhammad ibn Musa al-Khwarizmi, around 820 CE, the same mathematician behind the word algebra in Epistle 4, taught how to solve quadratic equations by completing the square. His method was geometric. He drew a square of side x, added rectangles for the x terms, and filled in the missing corner to complete a bigger square. It was the same completing the square you will see in the proof below, done with drawings instead of symbols. He only accepted positive answers, because in his time an equation described a real length or area.

The writing we use today came later, in two steps. François Viète, in 1591, was the first to use letters systematically in algebra, and he saw the link between the roots of an equation and its coefficients: for x² + px + q = 0, the two roots add to −p and multiply to q. René Descartes, in his 1637 La Géométrie, the same book that gave us the superscript notation in Epistle 1, simplified the letters into the x, y and z you use today. With that writing, the quadratic formula could finally be written as one short line.

So when you solve a quadratic equation, you are using a Babylonian idea, finished by Brahmagupta, taught as a method by al-Khwarizmi, and written in the letters of Viète and Descartes.

The proof: why the quadratic formula works
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The quadratic formula is not a rule to memorize. It is completing the square, done once, in letters, so you never have to do it again.

Start from standard form, with a not zero:

ax² + bx + c = 0

Divide every term by a:

x² + (b/a)x + c/a = 0

Move the number term to the right:

x² + (b/a)x = −c/a

Now complete the square on the left. Half of b/a is b/2a, and its square is (b/2a)². Add it to both sides:

x² + (b/a)x + (b/2a)² = (b/2a)² − c/a

The left side is now a perfect square:

(x + b/2a)² = b²/4a² − c/a

Write the right side over the common denominator 4a²:

(x + b/2a)² = (b² − 4ac) / 4a²

Take the square root of both sides. Remember a square has two square roots, so the ± appears here:

x + b/2a = ±√(b² − 4ac) / 2a

Finally, subtract b/2a from both sides:

x = (−b ± √(b² − 4ac)) / 2a

That is the formula. Every step was just the balance rule from Epistle 4, used on letters instead of numbers. And the part under the square root, b² − 4ac, has a name of its own: the discriminant. It decides how many real roots the equation has. If b² − 4ac is greater than 0, there are two distinct real roots. If it equals 0, there is one repeated root. If it is less than 0, there is no real root, because you cannot take the square root of a negative number and get a real answer.

Key rules (for reference while you solve)
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Standard form: ax² + bx + c = 0, with a not zero.

Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.

Discriminant: b² − 4ac. Greater than 0 means two distinct real roots. Equal to 0 means one repeated root. Less than 0 means no real roots.

To factorize x² + px + q: find two numbers that add to p and multiply to q. Then x² + px + q = (x + first number)(x + second number).

Sum of the roots = −b/a. Product of the roots = c/a.

If a product equals zero, one of the factors is zero. If (x − 2)(x − 3) = 0, then x = 2 or x = 3.

Problem 1. Solve x² − 5x + 6 = 0 by factorization.

Solution:

Find two numbers that multiply to 6 and add to −5. They are −2 and −3.

So x² − 5x + 6 = (x − 2)(x − 3) = 0.

If a product is zero, one of the factors is zero.

x − 2 = 0, so x = 2. Or x − 3 = 0, so x = 3.

Check: 2² − 5 × 2 + 6 = 4 − 10 + 6 = 0, and 3² − 5 × 3 + 6 = 9 − 15 + 6 = 0. Both roots work.

Problem 2. Solve x² − 4x − 1 = 0, leaving your answers in surd form.

Solution:

This one does not factorize into whole numbers, so use the formula. Here a = 1, b = −4 and c = −1.

x = (−b ± √(b² − 4ac)) / 2a

x = (4 ± √((−4)² − 4 × 1 × (−1))) / 2

x = (4 ± √(16 + 4)) / 2

x = (4 ± √20) / 2

Simplify √20 = 2√5:

x = (4 ± 2√5) / 2 = 2 ± √5

So x = 2 + √5 or x = 2 − √5.

Check: 2 + √5 is about 4.236, and 4.236² − 4 × 4.236 − 1 is about 0. The other root checks the same way. Leave the answer in surd form, exactly as in Epistle 2, unless the question asks for a decimal.

Problem 3. The length of a rectangle is 3 cm more than its width. Its area is 70 cm². Find the width of the rectangle.

Solution:

A word problem first becomes an equation. Let the width be x cm.

Then the length is (x + 3) cm, and the area is width times length:

x(x + 3) = 70

x² + 3x = 70

x² + 3x − 70 = 0

Factorize. Two numbers that multiply to −70 and add to 3 are 10 and −7:

(x + 10)(x − 7) = 0

So x = −10 or x = 7.

A width cannot be negative, so reject x = −10. The width is 7 cm, and the length is 7 + 3 = 10 cm.

Check: 7 × 10 = 70, as given. The numbers are 7 cm and 10 cm.

Note: "Let the width be x" is part of the answer, as in Epistle 4. And rejecting the negative root is part of the answer too. The equation gives two roots, but the question only accepts one of them.

Examiner's tip
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Three traps catch most candidates in this topic.

First, never forget the ± in the formula. The formula gives two roots, and a question asking for "the roots" wants both of them. A single root is half an answer.

Second, get the signs of a, b and c right before you put them into the formula. Write the equation in standard form first, with everything on one side and zero on the other, and read a, b and c off that form (Problem 2). The most common error in the whole topic is a wrong sign in b² − 4ac.

Third, in word problems, check that your answer makes sense in the words of the question, not only in the equation (Problem 3). If a length comes out negative, or a number of people comes out as a fraction, the setup is right but that root is not an answer. That check costs only one look.

JAMB speed note: for multiple choice questions, use the sum and product of the roots before you solve. The two roots add to −b/a and multiply to c/a, so add the roots in each option and check them against −b/a. The correct option is usually the only one that fits both the sum and the product, so you can often spot it before doing the full working.

Audience problems: try these yourself
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Problem A. Solve x² − 9x + 20 = 0 by factorization.

Problem B. Solve x² + 6x + 2 = 0, leaving your answers in surd form.

https://ayzeroxa.com.ng/#/blog/preparation-for-any-mathematics-examination/epistle-5-quadratic-equations

I am sharing a free maths lesson I wrote for students preparing for WAEC, JAMB, NECO and GCE. It is the fourth part of my series, Preparation for Any Mathematics Examination. No payment, no sign up. Just read it.

This one covers algebraic expressions and linear equations, from the very basics to exam questions, with full working shown. It also explains why the balance method works, not just the rule to memorize, and where algebra came from.

EPISTLE 4: ALGEBRAIC EXPRESSIONS AND LINEAR EQUATIONS

Things You Would Learn
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In this epistle, you will learn:

• What an algebraic expression really is, in plain language.

• Where algebra came from, and the people behind it.

• Why the balance method works. The proof, not just the rule to memorize.

• How to collect like terms, expand brackets, and factorize.

• How to solve linear equations in exam format, with full working shown.

• How to turn a word problem into an equation, and check your answer.

• How to work faster, with fewer mistakes, and solve problems you have never seen before.

Start from the basics
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An algebraic expression is a combination of numbers, letters, and operations.

Take 3x + 5. The letter x is a variable. It stands for a number we do not know yet. The 3 in front of x is the coefficient. It says "3 lots of x." The parts separated by + or − are called terms: 3x is one term, 5 is another.

Like terms have the same letter and the same power. 3x and 7x are like terms, so they can be added: 3x + 7x = 10x. But 3x and 3x² are not like terms. One is x, the other is x multiplied by itself. So they cannot be added into a single term. They just sit side by side: 3x + 3x².

An equation says two expressions are equal. 3x + 5 = 20 is an equation. Solving it means finding the value of x that makes the two sides equal. Here, only x = 5 works, because 3 × 5 + 5 = 20.

An expression is a phrase. An equation is a sentence with an equal sign. The whole of algebra is learning to move from one to the other, and back again.

A brief history and the people behind it
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Algebra is over a thousand years old, and its name tells you where it came from.

Muhammad ibn Musa al-Khwarizmi, a Persian mathematician working in Baghdad around 820 CE, wrote a book called al-Kitab al-Mukhtasar fi Hisab al-Jabr wal-Muqabala, "The Compendious Book on Calculation by Completion and Balancing." The word "algebra" comes from al-jabr in that title, which means completion, or putting a broken thing back together. The book taught how to solve equations by moving terms from one side to the other and balancing them. Even the word "algorithm" comes from his name, Latinized as Algoritmi.

But al-Khwarizmi was not the first to solve equations.

Diophantus of Alexandria, around 250 CE, wrote Arithmetica, a book of problems about unknown numbers. His algebra was written out in words and abbreviations, with no equal sign and no letters as we use them, but the ideas were already there: let the unknown stand for a number, and work out what it must be.

The writing we use today came much later, in two steps.

François Viète, a French mathematician, published In Artem Analyticem Isagoge in 1591. He was the first to use letters systematically in algebra: vowels for unknown quantities, consonants for known ones. For the first time, an equation could be written as a general pattern, not just one worked example.

René Descartes, in his 1637 book La Géométrie, the same book that gave us the superscript notation for indices in Epistle 1, simplified Viète's system into the one you use today: x, y, and z for unknowns, and a, b, and c for known quantities. When you write 3x + 5 = 20, you are writing it Descartes' way.

One more name belongs here, because you write his symbol in every equation.

Robert Recorde, a Welsh mathematician, invented the equal sign in 1557, in his book The Whetstone of Witte. He chose two parallel lines of the same length, he wrote, "because no two things can be more equal." The sign stuck, and so did the idea behind it: the two sides of an equation are the same number, written two ways.

So when you balance an equation, you are using al-Khwarizmi's method, written in Viète's and Descartes' letters, under Recorde's equal sign.

The proof: why the balance method works
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Every student is told: "do the same thing to both sides." Almost nobody is shown why that keeps the equation true.

It works because of what an equation is.

The equation a = b says that a and b are the same number, written two ways. If you add the same number c to the same number twice, you get the same result twice. So a = b must give a + c = b + c. The same holds for subtracting c, multiplying by c, and dividing by c (as long as c is not zero).

So when you solve 3x + 5 = 20, every step below is just this rule, used once:

Subtract 5 from both sides:

3x + 5 − 5 = 20 − 5

3x = 15

Divide both sides by 3:

3x ÷ 3 = 15 ÷ 3

x = 5

At no point did the equation stop being true, because each side was changed by the same amount. That is the whole proof. "Move a term across the equal sign and change its sign" is only this rule in shorthand: subtracting 5 from both sides is what moves the +5 across and turns it into −5.

Check your answer by putting it back:

3 × 5 + 5 = 15 + 5 = 20. The two sides match, so x = 5 is correct.

Key rules (for reference while you solve)
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To collect like terms: add or subtract only terms with the same letter and power. 3x + 7x = 10x.

To expand a bracket: multiply the outside term by every term inside. a(b + c) = ab + ac.

To factorize: take out what is common. ab + ac = a(b + c).

To balance: do the same thing to both sides of the equation. Add, subtract, multiply, or divide both sides by the same number.

To solve: use balancing to get the unknown on its own on one side.

To check: substitute your answer back into the original equation. Both sides must come out equal.

Problem 1. Solve 5x − 7 = 3x + 9.

Solution:

The unknown is on both sides, so collect the x terms first.

Subtract 3x from both sides:

5x − 3x − 7 = 9

2x − 7 = 9

Now add 7 to both sides:

2x = 16

Divide both sides by 2:

x = 8

Check: 5 × 8 − 7 = 40 − 7 = 33, and 3 × 8 + 9 = 24 + 9 = 33. Both sides are 33, so x = 8 is correct.

Note: the x terms were moved to the left and the number to the right. You could move them the other way instead. The answer is the same, because both moves are just balancing.

Problem 2. Expand and simplify 3(2x − 5) + 4(x + 2).

Solution:

Expand the first bracket. Multiply 3 by every term inside:

3 × 2x = 6x and 3 × (−5) = −15

So 3(2x − 5) = 6x − 15

Expand the second bracket:

4 × x = 4x and 4 × 2 = 8

So 4(x + 2) = 4x + 8

Now add the two results, collecting like terms:

6x − 15 + 4x + 8 = 10x − 7

Note: the −15 and the +8 are like terms too. Both numbers with no letter, so they combine to −7, exactly like the x terms combine to 10x.

Problem 3. The sum of two consecutive whole numbers is 31. Find the numbers.

Solution:

A word problem first becomes an equation. Let the smaller number be x.

Then the next whole number is x + 1, because consecutive numbers follow one after the other.

Their sum is 31, so:

x + (x + 1) = 31

Collect like terms:

2x + 1 = 31

Subtract 1 from both sides:

2x = 30

Divide both sides by 2:

x = 15

So the smaller number is 15, and the next one is 15 + 1 = 16.

Check: 15 + 16 = 31, as given. The numbers are 15 and 16.

Note: "Let the smaller number be x" is part of the answer. Writing it down is what turns words into algebra. Examiners award marks for that setup on its own.

Examiner's tip
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Three traps catch most candidates in this topic.

First, never move a term across the equal sign without changing its sign. You are not really "moving" it. You are subtracting it from both sides, and the sign change is what that looks like (Problem 1). Keep the balance rule in mind and the sign follows.

Second, when you expand a bracket, multiply the outside term by EVERY term inside, including the negative ones. 3(2x − 5) is 6x − 15, not 6x − 5 (Problem 2). A missed negative sign is the most common lost mark in bracket questions.

Third, in word problems, write "Let the number be x" first, and check your answer in the words of the question, not only in the equation (Problem 3). If the numbers do not add up to 31, the setup is wrong. That check costs only one addition.

JAMB speed note: for multiple-choice questions, you can test an option instead of solving. Put each option in place of x and see which one balances the equation. The correct option is the only one that makes both sides equal, so you can often spot it before doing the full working.

Audience problems: try these yourself
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Problem A. Solve 4x + 9 = 2x + 21.

Problem B. Expand and simplify 2(3x − 4) − 5(x − 1).

https://ayzeroxa.com.ng/#/blog/preparation-for-any-mathematics-examination

Try the two problems at the end of the lesson and drop your answers in the replies. I will confirm the correct ones.