Quadratic Equations

1. What This Chapter Covers

The chapter opens on a school sports ground. The committee of Dhannur High School wants a Kho-Kho court of 29 m by 16 m laid inside a rectangular plot of area 558 m², leaving a strip of equal width all round for spectators. How wide is the strip, and is it enough?

Call the strip x metres wide. The plot is then (29 + 2x) long and (16 + 2x) broad, so its area is (29 + 2x)(16 + 2x).

Setting that equal to 558 gives 4x² + 90x + 464 = 558, so 4x² + 90x − 94 = 0, and dividing by 2, 2x² + 45x − 47 = 0. The left side is a quadratic polynomial and the right side is zero. That is the shape of everything in this chapter.

The index allots this chapter 12 periods in October. It runs from textbook page 105 to page 128, with four numbered exercises and an Optional Exercise.

2. A Second Situation, and the Definition

Rani has a square metal sheet. She cuts a square of side 9 cm from each corner and turns up the flaps to make an open box holding 144 cm³. What was the sheet?

If the sheet has side x, the box is 9 cm tall on a base of (x − 18) by (x − 18). So 9(x − 18)² = 144, giving (x − 18)² = 16 and x² − 36x + 308 = 0. The sheet was 22 cm square.

Both situations produce the same shape, and the book names it. A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0.

Writing the terms in descending order of degree gives the standard form, and y = p(x) = ax² + bx + c is called a quadratic function. The book lists where such functions show up: the path of a rocket, the shape of a satellite dish or a telescope mirror, the orbit of a satellite, the flight of a projectile, and the stopping distance of a braking vehicle.

3. Simplify First, Then Decide

The condition a ≠ 0 does more work than it looks. You cannot tell whether an equation is quadratic by glancing at it; you have to simplify first. Example-2 makes the point with four cases.

Take x(x + 1) + 8 = (x + 2)(x − 2). The left side is x² + x + 8 and the right is x² − 4. Both contain x², so the equation looks quadratic — but the x² terms cancel, leaving x + 12 = 0. It is linear.

Now take (x + 2)³ = x³ − 4. That looks cubic. Expanding the left gives x³ + 6x² + 12x + 8, the x³ terms cancel, and 6x² + 12x + 12 = 0 remains. It is quadratic.

The book's own Remark draws the moral: one equation looked quadratic and was not, the other looked cubic and was. Often we need to simplify the given equation before deciding whether it is quadratic.

A related subtlety sits in the Try This box on page 106, which asks about (2x + 1)(3x + 1) = b(x − 1)(x − 2). Expanding gives (6 − b)x² + (5 + 3b)x + (1 − 2b) = 0, which is quadratic for every value of b except b = 6, where the x² term vanishes and the equation collapses to linear.

4. Roots, and the First Method

A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. The book checks this concretely: putting x = 1 into 2x² − 3x + 1 gives 2 − 3 + 1 = 0, so 1 is a root.

The first method for finding roots is factorisation, and it rests on a single fact: if a product of two numbers is zero, at least one of them is zero. Factor the quadratic into two linear pieces, set each to zero, and read off the roots.

Example-3 factorises 2x² − 5x + 3 = 0 by splitting the middle term. We need p and q with p + q = b = −5 and p × q = a × c = 6. Listing the factor pairs of 6 as (1, 6), (−1, −6), (2, 3) and (−2, −3), only (−2, −3) has sum −5.

So 2x² − 5x + 3 becomes 2x² − 2x − 3x + 3, which groups as 2x(x − 1) − 3(x − 1) = (2x − 3)(x − 1). Setting each factor to zero gives x = 3/2 and x = 1.

Example-5 returns to the Kho-Kho court. Factorising 2x² + 45x − 47 = 0 as (x − 1)(2x + 47) = 0 gives x = 1 or x = −47/2. A width cannot be negative, so the strip is 1 m wide — and the book's verdict is blunt: so it is not enough for spectators.

That rejection step matters as much as the algebra. A quadratic hands you two roots; the situation decides how many of them are answers.

The book's Do This box drills the sign pattern on four near-identical equations: x² + 5x + 6, x² − 5x + 6, x² + 5x − 6 and x² − 5x − 6. The first two factorise with both numbers the same sign, the last two with opposite signs, and the roots move from −2 and −3 to 2 and 3 to −6 and 1 to 6 and −1.

Reading the pattern is worth more than memorising it. When c is positive the two numbers share a sign, and that sign is the sign of b. When c is negative they have opposite signs, and the larger of the two carries the sign of b.

5. When Factorisation Fails

Try to factorise x² + 4x − 4 = 0. You need two numbers with sum 4 and product −4, and no integers do that. The method has run out.

The book's escape is to make the left side a perfect square on purpose. Since x² + 4x is x² + 2·x·2, adding 2² = 4 completes it.

Adding 4 to both sides of x² + 4x = 4 gives (x + 2)² = 8, so x + 2 = ±√8 and x = −2 ± 2√2. No factor pair was needed.

When the coefficient of x² is not 1, divide through first. For 3x² − 5x + 2 = 0 the book divides by 3, rewrites the middle term as 2·x·(5/6), adds (5/6)² = 25/36 to both sides, and lands on (x − 5/6)² = 1/36, giving x = 1 or x = 2/3.

The book states the procedure as a five-step algorithm: divide by a; move c/a to the right; add the square of half the coefficient of x to both sides; write the left as a square; and solve.

Example-7 shows the method detecting failure rather than producing it. For 4x² + 3x + 5 = 0 the steps lead to (x + 3/8)² = −71/64. A square of a real number cannot be negative, so the equation has no real roots — the method reports this cleanly instead of grinding to a halt.

6. The Formula, and Where It Comes From

Running the same five steps on the general equation ax² + bx + c = 0 is what produces the quadratic formula. Dividing by a, moving c/a across and adding the square of half of b/a gives

(x + b/2a)² = (b² − 4ac) / 4a²

and taking square roots, provided b² − 4ac is not negative, yields the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

The formula is not a separate technique. It is completing the square done once, in general, so that nobody has to do it again.

Example-8 applies it to the rectangular plot of area 528 m² whose length is one metre more than twice its breadth. With breadth x, the equation is 2x² + x − 528 = 0, so a = 2, b = 1 and c = −528. The discriminant is 1 + 4224 = 4225, whose square root is 65, giving x = 16 or x = −33/2. The breadth is 16 m and the length 33 m.

Example-13 runs the formula on a boat problem. A motor boat doing 18 km/h in still water takes 1 hour longer to go 24 km upstream than to return. With the stream at x km/h, 24/(18 − x) − 24/(18 + x) = 1 reduces to x² + 48x − 324 = 0, and the roots 6 and −54 leave 6 km/h as the only sensible speed.

7. The Discriminant

The quantity under the square root does all the deciding. Because b² − 4ac determines whether real roots exist, it is called the discriminant.

So a quadratic equation has two distinct real roots if b² − 4ac > 0, two equal real roots if b² − 4ac = 0, and no real roots if b² − 4ac < 0.

The geometry behind this is the link back to chapter 3. Roots of a quadratic equation are the points where the curve of the quadratic polynomial meets the X-axis, so a positive discriminant means the curve cuts the axis twice, zero means it touches once, and negative means it never reaches the axis.

Example-14 needs only the discriminant: for 2x² − 4x + 3 = 0 it is 16 − 24 = −8, so there are no real roots and no further work is required. Example-16 does the same for 3x² − 2x + 1/3 = 0, where 4 − 4 = 0 signals two equal roots, both equal to −b/2a = 1/3.

Example-15 is the best use of the idea. A pole must stand on the boundary of a circular park of diameter 13 m so that its distances from two diametrically opposite gates differ by 7 m. Is it even possible?

Because AB is a diameter, the angle at P is a right angle, so x² + (x + 7)² = 169, which reduces to x² + 7x − 60 = 0. Its discriminant is 289, comfortably positive, so the pole can be erected — at 5 m from one gate and 12 m from the other.

Exercise 5.4 turns the question round and asks whether certain designs are possible at all. A rectangular park of perimeter 80 m and area 400 m² gives (l − 20)² = 0, so it exists but is forced to be a square of side 20 m. Two friends whose ages sum to 20 and whose ages four years ago multiplied to 48 give a discriminant of −48, so no such pair exists.

8. Word Problems, and the Rejected Root

Example-10 sets a rectangular park against a triangular one. The rectangle's breadth is 3 m less than its length, and its area is 4 m² more than an isosceles triangle whose base is that breadth and whose altitude is 12 m.

With breadth x, the condition x² + 3x = 6x + 4 becomes x² − 3x − 4 = 0, whose roots are 4 and −1. A breadth cannot be −1, so the park is 4 m by 7 m, and the book verifies it: 28 − 24 = 4.

That rejection is the pattern across the whole chapter. Example-9 finds two consecutive positive odd integers whose squares sum to 290; the roots are 11 and −13, and only 11 survives the word positive, giving 11 and 13. Example-13 discards −54 km/h. Example-5 discards −47/2 metres.

A quadratic always offers two roots. The problem decides how many are answers, and the check is against the words, not the algebra.

9. Equations That Become Quadratic

Several problems in the chapter are not quadratic as written. They become quadratic once the fractions are cleared, and the clearing step carries a condition that must be stated.

Example-4 solves x − 1/(3x) = 1/6, noting x ≠ 0. Multiplying through by 6x gives 6x² − x − 2 = 0, which factorises as (3x − 2)(2x + 1) = 0, so the roots are 2/3 and −1/2.

Example-12 does two more. From x + 1/x = 3 comes x² − 3x + 1 = 0, whose discriminant is 5, giving roots (3 ± √5)/2. From 1/x − 1/(x − 2) = 3, with x ≠ 0 and x ≠ 2, multiplying by x(x − 2) gives 3x² − 6x + 2 = 0 and roots (3 ± √3)/3.

Those excluded values are not decoration. They are the numbers that would make a denominator zero, so they can never be roots however the algebra turns out, and a root that coincides with one of them must be discarded.

Exercise 5.3 leans on this repeatedly. Rehman's age comes from the reciprocals 1/(x − 3) + 1/(x + 5) = 1/3, which reduces to x² − 4x − 21 = 0 and gives 7 years. Two taps filling a tank in 9 and 3/8 hours, one 10 hours faster than the other, reduce to 4x² − 115x + 375 = 0 and give 25 hours and 15 hours.

Speed problems behave the same way. A train covering 360 km that would save an hour at 5 km/h more gives x² + 5x − 1800 = 0 and a speed of 40 km/h; the express and passenger trains over 132 km give x² + 11x − 1452 = 0, so 33 km/h and 44 km/h.

The Optional Exercise pushes further. Counting line segments between n points gives n(n − 1)/2 = 10 and so n = 5. A polygon's diagonals give n(n − 3)/2 = 65 and n = 13 — and asking the same of 50 diagonals produces n² − 3n − 100 = 0, whose discriminant 409 is not a perfect square, so no polygon has exactly 50 diagonals.

10. What the Exercises Ask, and Where the Book Goes Wrong

Exercise 5.1 has two questions: eight equations to test for quadraticness after simplification, and four situations to convert into quadratic equations. Exercise 5.2 has ten, starting with nine factorisations and running through numbers, triangles, a cottage industry, rectangles, two trains and a motor boat. Exercise 5.3 has thirteen, mostly formula work and word problems. Exercise 5.4 has five on the nature of roots, and the Optional Exercise adds seven.

Answers are printed at textbook pages 376 to 378. Exercise 5.4 is entirely correct, and so is nearly all of Exercise 5.2. But two printed answers are wrong and two more are careless.

Exercise 5.3 question 12 throws an object upward from a 12 m building, its height after t seconds being S = 12 + 17t − 5t². The key answers 3 seconds. Substituting t = 3 gives S = 12 + 51 − 45 = 18, so the object is still 18 m up.

Setting S = 0 gives 5t² − 17t − 12 = 0, whose discriminant is 529 and whose positive root is t = 4 seconds. And S(4) = 12 + 68 − 80 = 0 exactly.

Exercise 5.3 question 7 asks for two numbers whose squares differ by 180, the smaller one's square being 8 times the larger number. The key prints "18, 12; −18, −12". The second pair is impossible: if the larger number is negative, then s² = 8L asks for a square equal to a negative number. Only 18 works as the larger, giving the smaller as 12 or −12.

Two smaller slips. Exercise 5.2 question 3 asks for two consecutive positive integers whose squares sum to 613, and the key offers "17, 18; −17, −18" — the negative pair contradicts the question's own word. And Exercise 5.2 question 8 answers a question about speeds with "15 km, 20 km"; the values are right but the units should be km/h.

11. Four Things in the Chapter Text

One error here is worth more attention than all the answer-key slips together.

Page 119 prints the quadratic formula with denominator 2 instead of 2a. The line reads "Using the quadratic formula x = (−b ± √(b² − 4ac)) / 2", and the working that follows is correct only because that example has a = 1.

That is what makes it dangerous. A student who copies the formula from this line gets the right answer on that problem and the wrong answer on every equation where a ≠ 1. The chapter summary on page 128 states it correctly with 2a, which confirms page 119 is the misprint.

Three more, all minor. Page 128 opens its summary with "Standard form of quadrat2ic equation", a stray digit inside the word. Page 126 says the value of x gives "the distance from gate B to pole B", where P is meant. And page 120 prints a doubled equals sign in "= = 4 or −1".

Finally, a gap rather than an error. All nine parabolas drawn on page 124 for the three discriminant cases open upward, so a reader could conclude that every quadratic curve does.

The book knows better: its own Suggested Project on page 128 asks for graphs "for different situations like a > 0, a < 0, b = 0". But the a < 0 picture is never shown, and chapter 3 had already established that a negative leading coefficient flips the curve.

12. Summary

A quadratic equation is ax² + bx + c = 0 with a ≠ 0, and you cannot tell whether an equation qualifies until you have simplified it. Terms that look decisive often cancel, in both directions.

A root is a value that makes the left side zero. Factorisation finds roots by splitting the expression into linear factors and setting each to zero, which works whenever a convenient factor pair exists.

Completing the square works always. Divide by a, move the constant across, add the square of half the coefficient of x, and take square roots. Running those steps on the general equation is exactly what produces the quadratic formula, so the formula is not a separate trick but a shortcut through work already done.

The discriminant b² − 4ac settles the outcome before any root is computed: positive gives two distinct real roots and a curve cutting the axis twice, zero gives two equal roots and a curve touching it, negative gives no real roots and a curve that never reaches it.

Word problems end with a decision the algebra cannot make. A quadratic supplies two roots; lengths, speeds, ages and counts cannot be negative, so check each root against the words of the problem and say which one you are discarding and why.