Algebra

The first type is plain handling of expressions: multiply, expand, simplify and put the given values. These are the easiest marks of the chapter, provided you are careful with signs.

Expanding with identities. (3x − 5)² is not 9x² − 25. It is (3x)² − 2 × 3x × 5 + 5² = 9x² − 30x + 25. The middle term 2ab is exactly where marks are lost, so always write a and b first: here a = 3x and b = 5.

[[fig:0]]

Sum and difference of two squares. Two results come up again and again:

• (a + b)² + (a − b)² = 2(a² + b²)
• (a + b)² − (a − b)² = 4ab

Why? Add the two expansions and the 2ab and −2ab cancel, leaving a² + b² twice. Subtract them and a² and b² cancel, leaving 2ab − (−2ab) = 4ab. So (2x + 3y)² − (2x − 3y)² = 4 × 2x × 3y = 24xy, without any expansion.

[[fig:1]]

Substitution. When values are given, simplify first and put the values afterwards. Also look for a pattern. If x = 9 and the expression has the coefficient 10, then 10 = x + 1, and writing the coefficients in terms of x often makes a monster expression collapse. With x = 9, x³ − 10x² + 10x = x³ − (x + 1)x² + 10x = −x² + 10x = x(10 − x) = 9.

Comparing coefficients. If (2x + 1)(x + 3) = Ax² + Bx + C for all x, expand and match: 2x² + 7x + 3, so A = 2, B = 7, C = 3. There is a useful shortcut here. A + B + C is the value of the expression at x = 1: (2 + 1)(1 + 3) = 12 = 2 + 7 + 3.

Radhika simplifies (x + 4)² − (x − 4)². By the second result it is 4 × x × 4 = 16x, and a check at x = 1 gives 25 − 9 = 16.

This is the algebra question you will meet most often. Three quantities are tied together: the sum (a + b) or the difference (a − b), the product ab, and the sum of squares a² + b². The question gives two of them and asks the third.

Two identities say it all:

Formula (Sum of squares):
>> - a² + b² = (a + b)² − 2ab
>> - a² + b² = (a − b)² + 2ab

[[fig:0]]

From these follow two more results that link the sum and the difference directly:

• (a + b)² = (a − b)² + 4ab
• (a − b)² = (a + b)² − 4ab

[[fig:1]]

Say Radhika is told a + b = 9 and ab = 14. Then a² + b² = 81 − 28 = 53, and (a − b)² = 81 − 56 = 25, so a − b = ±5. The numbers are in fact 7 and 2.

Scaled forms. Very often the letters come with coefficients: "2x + 3y = 13, xy = 6, find 4x² + 9y²". Treat 2x as one number A and 3y as another number B. Then A + B = 13 and AB = 6xy = 36, so A² + B² = 169 − 72 = 97. The only care needed is that the product of the two "numbers" is 2x × 3y = 6xy, not xy.

Up to the fourth power. a⁴ + b⁴ = (a² + b²)² − 2a²b². Another very useful one:
a⁴ + a²b² + b⁴ = (a² + b²)² − a²b² = (a² + ab + b²)(a² − ab + b²).

The ± sign. A square root gives two answers: if (a − b)² = 25, then a − b = 5 or −5. The question usually settles it with "a > b", "positive value", or by keeping only one of them in the options.

a² − b² = (a + b)(a − b). Expand the right side and see: a² − ab + ab − b² = a² − b². The middle terms cancel each other, which is why this identity is so clean.

[[fig:0]]

You will use it in three ways.

1. Sum and difference given. If a + b and a − b are given, a² − b² is just their product. Going one step further, a⁴ − b⁴ = (a² + b²)(a² − b²) = (a² + b²)(a + b)(a − b).

2. Number problems. "The squares of two numbers differ by 56 and the numbers differ by 4" means (a + b) × 4 = 56, so a + b = 14 and the numbers are 9 and 5. Whenever a word problem talks about a difference of squares, divide it by the difference (or the sum) of the numbers.

3. Chains of brackets. (a − b)(a + b) = a² − b². Multiply by (a² + b²) and you get a⁴ − b⁴; multiply by (a⁴ + b⁴) and you get a⁸ − b⁸. Every new bracket doubles the power. This is the chain product.

[[fig:1]]

The same identity speeds up calculation: 64² − 36² = (64 + 36)(64 − 36) = 100 × 28 = 2800. A series such as 20² − 19² + 18² − 17² + … becomes (20 + 19) + (18 + 17) + …, which is only the sum of the numbers.

Radhika multiplies (2p + 5)(2p − 5) and writes (2p)² − 5² = 4p² − 25 straight away. There is no middle term to worry about.

Now we go one power up. Multiply (a + b)² by (a + b) once more and you get
(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b).

[[fig:0]]

The second form is the one you will use most, because it contains the same three things as before: the sum, the product and the cubes. Read it backwards and you get the main formula of this type:

Formula (Sum and difference of cubes):
>> - a³ + b³ = (a + b)³ − 3ab(a + b)
>> - a³ − b³ = (a − b)³ + 3ab(a − b)

And the factor forms:

• a³ + b³ = (a + b)(a² − ab + b²)
• a³ − b³ = (a − b)(a² + ab + b²)

So cube questions are of three kinds.

1. Sum (or difference) and product given, cubes asked. a + b = 8, ab = 15 ⇒ a³ + b³ = 512 − 3 × 15 × 8 = 512 − 360 = 152. As a check, the numbers are 5 and 3, and 125 + 27 = 152.

2. Cubes given, product or sum asked. a − b = 3 and a³ − b³ = 117 ⇒ 117 = 27 + 9ab ⇒ ab = 10. Put the known values into the same identity and solve for the unknown.

3. Numerical simplification. a³ + b³a² − ab + b² = a + b and a³ − b³a² + ab + b² = a − b. When you see a three-term denominator like 67 × 67 − 67 × 33 + 33 × 33, the answer is 67 + 33. One care: the numbers in the numerator and the denominator must be the same. If the numerator has 0.7 and the denominator has 0.07, adjust the decimal places first.

[[fig:1]]

Learn the expansions with their signs too: (a − b)³ = a³ − 3a²b + 3ab² − b³. The signs alternate +, −, +, −, and the coefficients 1, 3, 3, 1 stay the same in every cube.

[[fig:2]]

For example, (x + 2)³ = x³ + 3 × x² × 2 + 3 × x × 4 + 8 = x³ + 6x² + 12x + 8.

In this type a perfect cube is hidden inside the question. Your job is to recognise it.

1. The hidden (a + b)³. a³ + b³ + 3ab(a + b) = (a + b)³. So if a + b = 5, then a³ + b³ + 15ab = (a + b)³ = 125, because 3ab(a + b) = 3ab × 5 = 15ab. Look at the coefficient of ab: it is always 3 × (the given sum). In the same way, if a − b = 4, then a³ − b³ − 12ab = (a − b)³ = 64.

[[fig:0]]

2. The hidden (p + 1)³. p³ + 3p² + 3p + 1 = (p + 1)³. So p³ + 3p² + 3p = (p + 1)³ − 1, and p(p² + 3p + 3) is the same thing. If p = 9, the value is 10³ − 1 = 999.

3. When a + b = 0. If a + b = 0, then b = −a and a³ + b³ = 0. The question usually writes this with brackets: "if m + n = 30, find (m − 18)³ + (n − 12)³". Add the brackets: (m − 18) + (n − 12) = m + n − 30 = 0, so the answer is 0 straight away.

[[fig:1]]

4. Terms with coefficients. 8x³ + 27y³ is (2x)³ + (3y)³. Treat A = 2x and B = 3y as two numbers. Then A³ + B³ = (A + B)³ − 3AB(A + B), where AB = 6xy. The same approach handles 64a³ − b³ = (4a)³ − b³.

[[fig:2]]

5. Comparing coefficients. Written out, (3 − 2x)³ is 27 − 54x + 36x² − 8x³. If the question writes the x² term as 3px², then 3p = 36 and p = 12. Expand carefully, then match term by term. Keep one more result in mind: (a + b)³ + (a − b)³ = 2a³ + 6ab² = 2a(a² + 3b²), because the terms with odd powers of b cancel.

Radhika sees 8x³ + y³ + 6xy with 2x + y = 1. Since 3 × 2x × y × (2x + y) = 6xy, the whole expression is (2x + y)³ = 1.

This family is a special case of Type 2, but it comes so often that it gets its own place. Take a = x and b = 1x. Their product is always ab = x × 1x = 1, and that single fact makes every formula of Type 2 very simple.

[[fig:0]]

(x + 1x)² = x² + 2 + 1x², so

Formula (x and 1/x (squares)):
>> - If x + 1x = k, then x² + 1x² = k² − 2.
>> - If x − 1x = k, then x² + 1x² = k² + 2.

One step further: x⁴ + 1x⁴ = (x² + 1x²)² − 2. So from x + 1x = 3 we get x² + 1x² = 7 and x⁴ + 1x⁴ = 47.

[[fig:1]]

Moving between plus and minus. (x + 1x)² − (x − 1x)² = 4. So if x + 1x = 6, then (x − 1x)² = 36 − 4 = 32 and x − 1x = ±4√2. Also, x² − 1x² = (x + 1x)(x − 1x).

Going backwards. If x² + 1x² = 62 and x > 0, then x + 1x = √(62 + 2) = 8, and x − 1x = ±√(62 − 2) = ±2√15 (take the sign the question allows). Practise this reverse reading well, because questions often give the bigger power and ask the smaller one.

The root form. If √x + 1√x = k, square it: x + 2 + 1x = k², so x + 1x = k² − 2. It is the same formula, one level lower.

The hidden form: a quadratic. x² − 6x + 1 = 0 looks like an equation, but divide every term by x: x − 6 + 1x = 0, so x + 1x = 6. In the same way, x² − 6x − 1 = 0 gives x − 1x = 6. Whenever the first and last coefficients are equal in size (1 and 1), think of dividing by x.

[[fig:2]]

Parts with coefficients. In 3x + 13x or 2a + 5a, the two parts still have a fixed product (3x × 13x = 1, and 2a × 5a = 10). So (2a + 5a)² = 4a² + 25a² + 2 × 10. Find the product of the two parts first; that product decides the "2" in the formula.

The shifted form. In "m + 1m − 3 = 5", subtract 3 from both sides: (m − 3) + 1m − 3 = 2. Now t = m − 3 plays the role of x.

Take Radhika's case: x + 1x = 4. Then x² + 1x² = 14, x⁴ + 1x⁴ = 196 − 2 = 194, and (x − 1x)² = 16 − 4 = 12.

Now the cubes. With a = x, b = 1x and ab = 1, the cube identity (a + b)³ = a³ + b³ + 3ab(a + b) gives

Formula (x and 1/x (cubes)):
>> - If x + 1x = k, then x³ + 1x³ = k³ − 3k.
>> - If x − 1x = k, then x³ − 1x³ = k³ + 3k.

So if x + 1x = 4, then x³ + 1x³ = 64 − 12 = 52.

Higher powers are built from smaller ones. Multiply two known sums and remove the extra part:
(x² + 1x²)(x³ + 1x³) = x⁵ + 1x⁵ + x + 1x.
So x⁵ + 1x⁵ = (x² + 1x²)(x³ + 1x³) − (x + 1x).
And x⁶ + 1x⁶ = (x³ + 1x³)² − 2, or (x² + 1x²)³ − 3(x² + 1x²).
The second route works for the minus form as well, because x² + 1x² is a plus-sum in both cases.

[[fig:0]]

Four special values to learn.

• x + 1x = 2 ⇒ x = 1. Every power gives xⁿ + 1xⁿ = 2.
• x + 1x = −2 ⇒ x = −1. Then xⁿ + x⁻ⁿ = 2 for even n and −2 for odd n.
• x + 1x = −1 ⇒ x² + x + 1 = 0 ⇒ x³ = 1. Powers repeat in a cycle of 3: x⁴ = x, x⁵ = x², x⁶ = 1.
• x + 1x = 1 ⇒ x² − x + 1 = 0 ⇒ x³ = −1. So x⁶ = 1, and powers repeat in a cycle of 6.

[[fig:1]]

Why is x³ = 1 in the third case? Multiply x² + x + 1 = 0 by (x − 1) and you get x³ − 1 = 0. That one line opens many tough-looking questions such as x¹⁰⁰ + x¹⁰¹ + x¹⁰².

From a quadratic to a big power. If x² − 7x + 1 = 0, then x + 1x = 7 and x² + 1x² = 47. So x⁴ + 1 = 47x², that is, x⁴ − 47x² = −1. A question may ask x⁴ − 47x² + 5, and the answer is −1 + 5 = 4. Try to make the expression look like x²ⁿ + 1 = (value) × xⁿ.

With three letters the method is exactly the same as with two. Write (a + b + c)² = (a + b + c)(a + b + c) and multiply: every letter meets itself once (a², b², c²) and meets every other letter twice (2ab, 2bc, 2ca).

Formula (Square of three terms):
>> (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

[[fig:0]]

So once again three quantities are linked: the sum a + b + c, the sum of squares a² + b² + c², and the sum of pair-products ab + bc + ca. Any two give the third. If a + b + c = 10 and ab + bc + ca = 31, then a² + b² + c² = 100 − 62 = 38.

The second expression. a² + b² + c² − ab − bc − ca appears in many questions. Multiply it by 2 and rearrange:
2(a² + b² + c² − ab − bc − ca) = (a − b)² + (b − c)² + (c − a)².
So a² + b² + c² − ab − bc − ca = 12[(a − b)² + (b − c)² + (c − a)²].

This form is a real help when a, b, c are close to each other. Radhika takes a = 51, b = 52, c = 53: the differences are 1, 1 and 2, so the value is 12(1 + 1 + 4) = 3. Any three consecutive numbers give 3, and numbers in AP with common difference d give 12(d² + d² + 4d²) = 3d².

[[fig:1]]

Sign changes. If the question has a + b − c, replace c by (−c) everywhere:
(a + b − c)² = a² + b² + c² + 2(ab − bc − ca).

Recognising a perfect square of three terms. In a² + 9b² + 4c² − 6ab + 12bc − 4ca, the square roots of the square terms are a, 3b and 2c. Now fix the signs from the cross terms: −6ab means a and 3b have opposite signs, and +12bc means 3b and 2c have the same sign. So it is (a − 3b − 2c)². The last cross term confirms it: 2 × a × (−2c) = −4ca.

This is the best-known three-variable identity:

Formula (Three cubes):
>> a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

You can verify it by multiplying out the right side: all the mixed terms like a²b cancel and only a³ + b³ + c³ − 3abc remains. For the exam, what matters is that it can be written in three useful ways. Let S = a + b + c, Q = a² + b² + c² and P = ab + bc + ca.

• Face 1: a³ + b³ + c³ − 3abc = S(Q − P)
• Face 2: = S(S² − 3P), because Q = S² − 2P
• Face 3: = 12S[(a − b)² + (b − c)² + (c − a)²]

Choose the face that matches the given data. If S and P are given, use Face 2. If S and Q are given, find P first. If the actual numbers are given and they are close to each other, use Face 3.

[[fig:0]]

Radhika takes a = 31, b = 32, c = 33. Here S = 96 and the bracket of Face 3 is 1 + 1 + 4 = 6, so a³ + b³ + c³ − 3abc = 12 × 96 × 6 = 288, and not a single cube had to be worked out.

[[fig:1]]

Finding a³ + b³ + c³ itself. Move 3abc to the other side:
a³ + b³ + c³ = S(Q − P) + 3abc.
If a³ + b³ + c³ is given and abc is asked, the same equation gives abc.

Terms with coefficients. 27x³ + 8y³ + 125z³ − 90xyz is (3x)³ + (2y)³ + (5z)³ − 3(3x)(2y)(5z). So treat 3x, 2y and 5z as the three numbers a, b, c.

Look at the big identity again: a³ + b³ + c³ − 3abc = (a + b + c)(…). If a + b + c = 0, the right side becomes 0, so

Theorem (Sum zero):
>> a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc.

[[fig:0]]

The whole skill here is to spot that the sum is zero, because the question keeps it hidden.

• Numbers: in 47³ − 30³ − 17³, 47 + (−30) + (−17) = 0, so the value is 3 × 47 × (−30) × (−17). Decimals work the same way: 2.35 + 1.15 − 3.5 = 0.
• Brackets: (x − y) + (y − z) + (z − x) = 0 always. So (x − y)³ + (y − z)³ + (z − x)³ = 3(x − y)(y − z)(z − x), with no condition at all.
• Hidden constants: "y = 3x − 2, find 27x³ − y³ − 8 − 18xy". Write 3x − y − 2 = 0. The three numbers are 3x, −y and −2. Their cubes are 27x³, −y³ and −8, and 3 × (3x)(−y)(−2) = 18xy. So 27x³ − y³ − 8 = 18xy and the expression is 0.

[[fig:1]]

Other results when a + b + c = 0.

• a + b = −c, so (a + b)² = c², that is, a² + b² + 2ab = c².
• a² + b² + c² = −2(ab + bc + ca).
• a²bc + b²ca + c²ab = a³ + b³ + c³abc = 3.

Putting values works perfectly here. Take a = 1, b = 1, c = −2 (or a = 1, b = −1, c = 0, provided no denominator becomes zero). An expression with a fixed value shows that value at once.

Radhika checks the rule with a = 2, b = 3, c = −5: a³ + b³ + c³ = 8 + 27 − 125 = −90, and 3abc = 3 × 2 × 3 × (−5) = −90.

The equation form. If A³ + B³ + C³ = 3ABC for real numbers, then either A + B + C = 0 or A = B = C. Many questions give such an equation with brackets in x and ask x; try A + B + C = 0 first.

The square of a real number is never negative: (anything)² ≥ 0. So if several squares add up to zero, there is only one possibility: every square must be zero.

[[fig:0]]

(x − 3)² + (y + 1)² = 0 ⇒ x − 3 = 0 and y + 1 = 0 ⇒ x = 3, y = −1.

The question never gives the squares directly. It gives something like x² + y² − 6x + 2y + 10 = 0, and you have to complete the squares:

• x² − 6x = (x − 3)² − 9 (take half the coefficient of x, that is 3, and square it)
• y² + 2y = (y + 1)² − 1

The equation becomes (x − 3)² + (y + 1)² − 10 + 10 = 0, that is, (x − 3)² + (y + 1)² = 0. The constant 10 was chosen so that everything cancels.

[[fig:1]]

The quick way. For x² − 2kx you need the square (x − k)². So in a question like a² + b² + c² = 2(4a − 3b + 5c) − 50, move everything to one side: a² − 8a + b² + 6b + c² − 10c + 50 = 0. Each value is half its coefficient with the opposite sign, so a = 4, b = −3, c = 5, and the check is 16 + 9 + 25 = 50. If the check works, the answer is correct.

Squares with coefficients. 4x² − 12x is (2x)² − 2 × 2x × 3, so it belongs to (2x − 3)² = 4x² − 12x + 9. Then 2x = 3 and x = 32.

The other zero sum: a = b = c. If a² + b² + c² = ab + bc + ca, then 12[(a − b)² + (b − c)² + (c − a)²] = 0, so a = b = c. After that, every expression reduces to a ratio of coefficients.

For example, take x² + 4y² − 2x + 4y + 2 = 0. Here (x − 1)² = x² − 2x + 1 and (2y + 1)² = 4y² + 4y + 1, and the constants 1 + 1 = 2 match. So x = 1 and y = −12.

A linear equation has its variables only in the first power: 3x + 5 = 20, or 2x + 3y = 12. Solving it means finding the value that makes both sides equal. The rule never changes: whatever you do to one side, do to the other side as well.

[[fig:0]]

One variable. 3x + 5 = 20 ⇒ 3x = 15 ⇒ x = 5. With fractions, multiply by the LCM of the denominators first: x3 + x4 = 14 ⇒ 4x + 3x = 168 ⇒ x = 24.

Word problems. Read the story and turn every sentence into an equation. "A number" = x, "5 more than it" = x + 5, "twice the number" = 2x, and "consecutive numbers" = x, x + 1, x + 2. For an odd count of consecutive numbers, take the middle one as x: five consecutive numbers are x − 2, x − 1, x, x + 1, x + 2, and their sum is 5x.

Take Radhika's notes. She has ₹10 and ₹20 notes, 25 notes in all, worth ₹380. Let the number of ₹20 notes be n. Then 20n + 10(25 − n) = 380 ⇒ 10n = 130 ⇒ n = 13, so she has 13 notes of ₹20 and 12 notes of ₹10.

This is the "heads and legs" way of thinking: assume everything is of one kind, see how much is short, and divide by the difference per item. If all 25 notes were of ₹10, the total would be ₹250. The shortage is ₹130 and each ₹20 note adds ₹10 more, so 13 notes are of ₹20.

[[fig:1]]

Two variables (a system of equations). There are two methods:

• Elimination: make the coefficient of one variable equal, then add or subtract. From 3x + 2y = 16 and x − 2y = 0, adding gives 4x = 16, so x = 4 and y = 2.
• Substitution: find one variable from one equation and put it in the other. If y = 2x − 1 and 3x + y = 9, then 3x + 2x − 1 = 9 ⇒ x = 2, y = 3.

[[fig:2]]

Three variables. Add or subtract the equations to remove one variable at a time. Often one well-chosen step is enough: if x + y + z = 10 and x + y − z = 4, subtracting gives 2z = 6.

Checking the options saves a lot of work here. When the options are complete solutions (like x = 2, y = 3, z = 1), put each option into the equations. The option that satisfies every equation is the answer.

Equations that can be made linear. 4x + 3y = 2xy is not linear as it stands, but divide by xy: 4y + 3x = 2. Now 1x and 1y act as the new variables.

Every linear equation in two variables, like 2x + 3y = 6, is a straight line on a graph. Two such equations are two lines, and two lines can behave in only three ways:

• They cut at one point ⇒ exactly one (unique) solution.
• They are parallel and never meet ⇒ no solution.
• They are the same line, one lying on the other ⇒ infinitely many solutions.

[[fig:0]]

To decide which case it is, compare the ratios of the coefficients. For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:

[[fig:1]]

Why does this work? If a₁a₂ = b₁b₂, the x- and y-parts of the two equations are multiples of each other. Then either the constants follow the same multiple (the same line) or they do not (parallel lines, which contradict each other).

Radhika compares x + 2y = 3 and 2x + 4y = 7. The ratios are 12 = 24, but 37 is different, so the lines are parallel and there is no solution. Had the second equation been 2x + 4y = 6, all three ratios would be 12 and there would be infinitely many solutions.

Two cautions. Bring both equations to the same form before taking ratios (all terms on one side in both, or the constants on the right in both). And when a question says "parallel lines", it means the same thing as "no solution".

A polynomial in x is an expression like 3x³ − 2x² + x − 7, in which every power of x is a whole number (0, 1, 2, 3, …). So √x, 1x or x⁻² stop an expression from being a polynomial. The highest power is the degree: degree 1 is linear, 2 is quadratic and 3 is cubic.

Factorisation means writing a polynomial as a product. Try these tools in this order:

• Take out the common factor: 3x² − 27 = 3(x² − 9).
• Use an identity: 3(x² − 9) = 3(x + 3)(x − 3). Also a² ± 2ab + b² = (a ± b)².
• Group the terms: ab + 2a + 3b + 6 = a(b + 2) + 3(b + 2) = (a + 3)(b + 2).
• Look for hidden squares: x² − y² + 4y − 4 = x² − (y − 2)² = (x + y − 2)(x − y + 2).
• Split the middle term: for x² + 5x + 6, find two numbers with sum 5 and product 6 (2 and 3), so it is (x + 2)(x + 3).

[[fig:0]]

Remainder theorem. When a polynomial p(x) is divided by (x − a), the remainder is p(a). Write p(x) = (x − a) × q(x) + R and put x = a: the first part becomes 0, so p(a) = R. This is why a remainder never needs long division.

[[fig:1]]

For a divisor like (2x − 3), put the value that makes it zero, x = 32. The remainder is p(32).

Factor theorem. If p(a) = 0, then (x − a) is a factor of p(x), because the remainder is zero. This finds unknown coefficients quickly: "if (x − 1) is a factor of x² + kx − 4, find k" gives 1 + k − 4 = 0, so k = 3.

Radhika divides p(x) = x³ − 2x + 5 by (x + 2). She puts x = −2: −8 + 4 + 5 = 1, so the remainder is 1. To make the polynomial exactly divisible, she must subtract 1 (that is, add −1).

If a polynomial is divisible by (x² − 1) = (x − 1)(x + 1), then p(1) = 0 and p(−1) = 0 both hold. That gives two conditions for two unknowns.

A quadratic equation is ax² + bx + c = 0 with a ≠ 0. It has at most two roots (solutions).

Solving.

• Factorisation: for x² − 7x + 12 = 0, the two numbers with sum −7 and product 12 are −3 and −4, so (x − 3)(x − 4) = 0 ⇒ x = 3 or 4.
• Formula: x = −b ± √b² − 4ac2a. It always works, even when factorisation is hard.

Formula (Quadratic formula):
>> x = −b ± √b² − 4ac2a

[[fig:0]]

In word problems a quadratic appears when a number is multiplied by itself or by a neighbour: "a number times 3 more than it is 40" ⇒ x(x + 3) = 40 ⇒ x = 5. Usually one root is negative or not a natural number, and the story rules it out.

Sum and product of roots. If α and β are the roots, then α + β = −ba and αβ = ca. So an equation with given roots is x² − (sum)x + (product) = 0. For a cubic ax³ + bx² + cx + d = 0 with roots α, β, γ, the sum is α + β + γ = −ba.

Nature of roots: the discriminant D = b² − 4ac.

Formula (Nature of roots, D = b² − 4ac):
>> - D > 0 ⇒ two real and distinct roots
>> - D = 0 ⇒ two real and equal roots (x = −b2a)
>> - D < 0 ⇒ no real roots

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Say Radhika wants equal roots for x² + 6x + k = 0. Then D = 36 − 4k = 0, so k = 9, and indeed x² + 6x + 9 = (x + 3)².

Least and greatest values. Completing the square shows them directly. x² − 6x + 14 = (x − 3)² + 5, and a square is never negative, so the least value is 5 (at x = 3). In general, for a > 0, ax² + bx + c has the least value 4ac − b²4a at x = −b2a; for a < 0 the same expression gives the greatest value.

For positive numbers, also remember AM ≥ GM (the arithmetic mean is never less than the geometric mean), with equality only when the numbers are equal. Two useful results follow:

• If the sum of positive numbers is fixed, their product is greatest when they are equal.
• For any positive x, y, z, (x + y + z)(1x + 1y + 1z) ≥ 9, with equality when x = y = z. So when the sum is fixed, the sum of the reciprocals is least when the numbers are equal.

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Comparing the roots of two equations. Some questions give Quantity I (an equation in x) and Quantity II (an equation in y) and ask for the relation between x and y. Solve both equations, write all the roots, and compare every root of one with every root of the other. If each root of one is greater than each root of the other, the relation is clear. If the answer changes from one pair to another, the relation cannot be established.

In Type 15 you saw that the roots of ax² + bx + c = 0 add up to −ba and multiply to ca. This type shows how much work those two numbers can do for you. Most of the time you never need the roots themselves.

Let the roots be α and β. Then α + β = −ba and αβ = ca.

Formula (Sum and product of roots):
>> α + β = −ba, αβ = ca

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Symmetric expressions. Any expression that stays the same when α and β are swapped can be written using only the sum and the product:

• α² + β² = (α + β)² − 2αβ
• (α − β)² = (α + β)² − 4αβ, and this equals Da², where D = b² − 4ac
• 1α + 1β = α + βαβ
• α³ + β³ = (α + β)³ − 3αβ(α + β)
• αβ + βα = α² + β²αβ

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Conditions on the roots. A condition on the roots turns into a condition on the coefficients.

• One root is zero: c = 0. The roots are equal in size and opposite in sign: b = 0.
• The roots are reciprocals of each other: αβ = 1, so a = c.
• One root is double the other: call the roots t and 2t. The sum 3t and the product 2t² give 2b² = 9ac.
• The roots differ by d: (α − β)² = d², so Da² = d².

Making a new equation. To build an equation whose roots are 2α and 2β, or α² and β², or α + 1 and β + 1, find the new sum S and the new product P in terms of α + β and αβ, and write x² − Sx + P = 0. Take Radhika's equation x² − 5x + 3 = 0 and suppose she wants roots that are one more than these. The new sum is (α + β) + 2 = 7 and the new product is αβ + (α + β) + 1 = 9, so the new equation is x² − 7x + 9 = 0.

Common root. If two equations share a root, that root satisfies both. Subtract one equation from the other to remove x² and you get the common root directly. Put it back into either equation to find the unknown.

Every quadratic expression ax² + bx + c has one turning point, called the vertex. Complete the square and you can see it:

Formula (Vertex form):
>> ax² + bx + c = a(x + b2a)² + 4ac − b²4a

The bracket is a square, so it is never negative. What happens next depends only on the sign of a.

• If a > 0, the square term can only add, so the expression is never below 4ac − b²4a. This is the least value.
• If a < 0, the square term can only subtract, so the expression is never above 4ac − b²4a. This is the greatest value.
• In both cases the value is reached at x = −b2a.

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Take x² − 8x + 3. Here a = 1 > 0, the vertex is at x = 4, and the least value is 16 − 32 + 3 = −13. For 9 + 2x − x² the sign of a is negative, so we get a greatest value: at x = 1 it is 9 + 2 − 1 = 10.

Two parts with a fixed sum. If x + y = s, then xy = x(s − x) is a quadratic with a < 0, so the product is greatest when x = y = s2. In the same way x² + y² = 2(x − s2)² + s²2 is least when x = y. This is the idea behind "greatest area for a fixed fence" questions.

Distances to several points. The sum (x − p)² + (x − q)² + (x − r)² is least when x is the mean of p, q and r.

A fraction with a fixed numerator. kax² + bx + c (k > 0, a > 0) is greatest where the denominator is least.

Whole-number restriction. If x must be a whole number and −b2a is not, test the whole numbers on either side of it. The real minimum or maximum at the vertex is then not allowed.

Stories. Revenue = price × number sold, area = length × breadth. Write the quantity in one variable, and it is almost always a quadratic with a < 0.

Some equations are not quadratic to look at, but a quadratic is hiding inside them. The skill is to spot the part that repeats, give it a new name, and solve the easy equation that results.

Even powers: put y = x². For x⁴ − 5x² + 4 = 0, note that x⁴ = (x²)². With y = x² the equation is y² − 5y + 4 = 0, so y = 1 or 4, and then x² = 1 or 4 gives x = ±1 or ±2. The same idea works for any equation whose powers of x are all even, and for any bracket that appears twice, such as (x² − 3x)² − 2(x² − 3x) − 8 = 0, where y = x² − 3x.

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x + 1x type. If an equation has x² + 1x² and x + 1x together, put y = x + 1x and use x² + 1x² = y² − 2. Remember that a real x makes x + 1x at least 2 or at most −2, so a value of y between −2 and 2 gives no real x.

Exponential type. In 9^x − 4 × 3^x + 3 = 0, put y = 3^x. Then 9^x = y². Since 3^x is always positive, drop any negative y.

Fractions. Multiply through by the common denominator and you get a quadratic. Then check that no root makes a denominator zero. For example, 1x + 1x + 2 = 34 becomes 4(2x + 2) = 3x(x + 2), that is 3x² − 2x − 8 = 0, with roots 2 and −43. Both are allowed here.

Square roots. In √f(x) = g(x), square both sides, solve, and then test each root in the original equation. A root with g(x) < 0 is extra and must be dropped, because a square root is never negative. For instance √x + 2 = x gives x² − x − 2 = 0, so x = 2 or −1, and only x = 2 works.

When a quadratic hides inside a longer product, such as (x + 1)(x + 2)(x + 3)(x + 4), pair the factors so that the two pairs share the same x-part (here x² + 5x), and put y equal to that part.

Almost every quadratic word problem works the same way. Name the unknown, translate the story into one equation, solve it, and then ask whether each root makes sense in the story.

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Numbers. Two consecutive numbers are x and x + 1 (consecutive even or odd numbers: x and x + 2). If their product is 156, then x(x + 1) = 156 gives x = 12, so the numbers are 12 and 13. The negative root x = −13 is a valid pair of numbers (−13 and −12) unless the question says positive or natural, so read the words carefully.

Geometry. Length × breadth gives the area, and 2(length + breadth) gives the perimeter. When the perimeter is known, length + breadth is known, and the area equation is x(s − x) = A. For a right-angled triangle use a² + b² = c² with the sides written as x and x + d.

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Speed and time. For a fixed distance d, the time at speed x is dx. If the speed rises by u, the time falls, so dx − dx + u = (time saved). Multiply through and you get a quadratic in x.

Work. If A alone takes x days and B alone takes x + k days, their one-day work is 1x + 1x + k, and this equals 1t when they finish together in t days.

Sharing. If n people share a cost C equally, each pays Cn. When 2 more join, each pays Cn + 2, and the difference gives the equation.

Ages. Write the two ages as x and x + d (or x and x − d), put the product or the sum of squares on one side, and keep only the positive root.

Radhika runs a stall and sells (100 − 2p) toffees a day at ₹p each. Her daily takings are p(100 − 2p), which is a quadratic with a < 0, so the greatest takings come at p = 25, exactly the vertex idea from Type 17.

An inequality compares two quantities with <, >, ≤ or ≥. Solving one is almost the same as solving an equation, with a single new rule.

The rules.

• Adding or subtracting the same number on both sides keeps the sign.
• Multiplying or dividing by a positive number keeps the sign.
• Multiplying or dividing by a negative number turns the sign round: from −2x < 6 we get x > −3.

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Compound inequalities. For −3 < 2x + 1 ≤ 9, do the same step on all three parts: −4 < 2x ≤ 8, so −2 < x ≤ 4. The integers that fit are −1, 0, 1, 2, 3 and 4. A strict sign (< or >) leaves the end out, and a sign with a bar (≤ or ≥) keeps it.

Modulus. |x| is the distance of x from 0, so it is never negative. In the same way |x − a| is the distance of x from a.

Formula (Modulus):
>> - |x − a| = b means x = a + b or x = a − b.
>> - |x − a| < b means x is within b of a: a − b < x < a + b (one piece).
>> - |x − a| > b means x is more than b away from a: x < a − b or x > a + b (two pieces).
>> - |x − a| = |x − b| means x is the same distance from both, so x = a + b2.

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Radhika solves |x − 1| < 3 by reading it as "x is within 3 of 1". The numbers run from −2 to 4, and the end points are not included.

Sum of distances. |x − a| + |x − b| is least when x lies between a and b, and the least value is the gap |a − b|. For three points, |x − p| + |x − q| + |x − r| is least at the middle point.

Quadratic inequalities. Factorise first. For p < q, (x − p)(x − q) < 0 holds between the roots, and (x − p)(x − q) > 0 holds outside them. For example x² − 7x + 10 ≤ 0 is (x − 2)(x − 5) ≤ 0, so 2 ≤ x ≤ 5.

Stories. "At most" gives ≤ and "at least" gives ≥. When the answer is a count of things, take the greatest or least whole number that fits.