Chapter 4 · Quadratic Equations

4.2 Solving Quadratic Equations by Factorisation

Write $ax^2+bx+c$ as a product of two linear factors and the equation splits in two: a product is zero only when one of its factors is zero. Splitting the middle term finds the factors.

In Section 4.1 we wrote the school-garden problem as $2x^2+3x-90=0$ and then simply checked that $x=6$ works. But where did $6$ come from? Guessing is slow, and even a lucky guess does not tell you whether there is another root. We need a method that finds all the roots, step by step.

The first method is factorisation. You already know how to factorise a quadratic polynomial by splitting the middle term (Section 2.2). Once the left side of the equation is written as a product of two brackets, a simple fact finishes the job: if two numbers multiply to give zero, one of them must be zero. For the garden, $2x^2+3x-90=(x-6)(2x+15)$, so $x=6$ or $x=-\tfrac{15}{2}$ — both roots, and no guessing.

The zero-product rule

📐 Theorem 4.2 — The zero-product rule
For real numbers $p$ and $q$: $\;pq=0$ if and only if $p=0$ or $q=0$ (or both).
Consequently, if $ax^2+bx+c=(px+q)(rx+s)$ with $p\ne0$ and $r\ne0$, then the roots of $ax^2+bx+c=0$ are $$x=-\frac qp\quad\text{and}\quad x=-\frac sr .$$

Why it is true. If $p=0$ or $q=0$, the product is clearly $0$. Conversely, suppose $pq=0$ and $p\ne0$. Dividing both sides by $p$ gives $q=\tfrac0p=0$. So at least one factor is zero. Applied to the equation $(px+q)(rx+s)=0$: either $px+q=0$, giving $x=-\tfrac qp$, or $rx+s=0$, giving $x=-\tfrac sr$.

✏️ Example 1 — A first factorisation
Solve $x^2-3x-10=0$.
Find the pair We need two numbers with sum $-3$ (the coefficient of $x$) and product $1\times(-10)=-10$. Factor pairs of $10$: $1\times10$ and $2\times5$. With opposite signs, $-5+2=-3$. ✓
Factorise $x^2-5x+2x-10=x(x-5)+2(x-5)=(x-5)(x+2)$.
Zero-product rule $(x-5)(x+2)=0$, so $x-5=0$ or $x+2=0$: $\;x=5$ or $x=-2$.
Check $25-15-10=0$ ✓ and $4+6-10=0$ ✓.
⚠️ Two classic mistakes

1. The right side must be zero. The rule works only for "product $=0$". From $(x-2)(x-3)=2$ you cannot conclude $x-2=2$ or $x-3=2$ (try $x=4$: $(2)(1)=2$ works, but $x=5$ gives $(3)(2)=6$). Expand first: $x^2-5x+6=2$, so $x^2-5x+4=0$, i.e. $(x-1)(x-4)=0$ and $x=1$ or $x=4$.

2. Never divide by $x$. Dividing $x^2=5x$ by $x$ gives only $x=5$ and loses the root $0$ (dividing by $x$ is not allowed when $x$ might be $0$). Instead bring everything to one side: $x^2-5x=0$, $\;x(x-5)=0$, so $x=0$ or $x=5$.

Interactive 1 — Two Factors, One Product

The two factors are drawn as straight lines, red and green; their product is the blue parabola. A tracer moves along the $x$-axis and shows the value of each factor and of the product. Watch: the parabola meets the $x$-axis exactly where one of the lines crosses it — the zero-product rule in action. Press Play, or drag the tracer.

1
first factor—
second factor—
product—

Splitting the middle term

Most quadratics do not arrive already factorised. Here is the method that turns $ax^2+bx+c$ into a product.

🧭 Problem-solving strategy — solving $ax^2+bx+c=0$ by factorisation
  1. Bring the equation to standard form $ax^2+bx+c=0$ (and divide by any common factor).
  2. Work out the product $ac$.
  3. Find two numbers $m$ and $n$ with $m+n=b$ and $mn=ac$. List the factor pairs of $|ac|$ and choose the signs:
    • if $ac\gt0$, both numbers have the same sign as $b$;
    • if $ac\lt0$, they have opposite signs, and the one that is larger in size has the sign of $b$.
  4. Split the middle term: $bx=mx+nx$. Group the first two and the last two terms and take out a common factor from each group; the same bracket must appear twice.
  5. Use the zero-product rule: set each factor equal to $0$ and solve.
  6. Check: substitute the roots, or use $\alpha+\beta=-\tfrac ba$ and $\alpha\beta=\tfrac ca$.
✏️ Example 2 — When $a\ne1$, and a repeated root
Solve (a) $6x^2-x-2=0$   (b) $9x^2-12x+4=0$.
(a) Pair $ac=6\times(-2)=-12$ and $b=-1$. Opposite signs, the larger negative: $-4+3=-1$ and $(-4)(3)=-12$. ✓
(a) Factorise $6x^2-4x+3x-2=2x(3x-2)+1(3x-2)=(2x+1)(3x-2)$.
(a) Roots $2x+1=0$ or $3x-2=0$: $\;x=-\tfrac12$ or $x=\tfrac23$. Check: sum $-\tfrac12+\tfrac23=\tfrac16=-\tfrac{-1}{6}$ ✓, product $-\tfrac13=\tfrac{-2}{6}$ ✓.
(b) $ac=36$, $b=-12$: both negative, $-6+(-6)=-12$. So $9x^2-6x-6x+4=3x(3x-2)-2(3x-2)=(3x-2)(3x-2)=(3x-2)^2$.
(b) Roots Both factors give $3x-2=0$, so $x=\tfrac23$ — the same root twice. We say the equation has a repeated root and write the roots as $\tfrac23,\tfrac23$. Its parabola just touches the $x$-axis at $x=\tfrac23$. Check: $9\cdot\tfrac49-12\cdot\tfrac23+4=4-8+4=0$ ✓.
✏️ Example 3 — Surds in the coefficients
Solve $\sqrt3\,x^2+10x+7\sqrt3=0$.
Pair $ac=\sqrt3\times7\sqrt3=21$ and $b=10$: we need sum $10$ and product $21$, i.e. $7$ and $3$.
Split $\sqrt3\,x^2+3x+7x+7\sqrt3$. In the first group write $3=\sqrt3\cdot\sqrt3$: $\;\sqrt3\,x(x+\sqrt3)+7(x+\sqrt3)=(\sqrt3\,x+7)(x+\sqrt3)$.
Roots $x=-\sqrt3$ or $x=-\dfrac{7}{\sqrt3}=-\dfrac{7\sqrt3}{3}$.
Check $x=-\sqrt3$: $\sqrt3\cdot3-10\sqrt3+7\sqrt3=0$ ✓. Product of roots $=\sqrt3\cdot\tfrac{7}{\sqrt3}=7=\tfrac{7\sqrt3}{\sqrt3}=\tfrac ca$ ✓.
✏️ Example 4 — Clearing fractions first
Solve $x+\dfrac1x=\dfrac52$, where $x\ne0$.
Clear fractions Multiply both sides by $2x$ (allowed, since $x\ne0$): $\;2x^2+2=5x$, so $2x^2-5x+2=0$.
Factorise $ac=4$, $b=-5$: $-4$ and $-1$. $\;2x^2-4x-x+2=2x(x-2)-1(x-2)=(2x-1)(x-2)$.
Roots $x=\tfrac12$ or $x=2$. Neither is $0$, so both are allowed.
Check $2+\tfrac12=\tfrac52$ ✓ and $\tfrac12+2=\tfrac52$ ✓ — a number and its reciprocal, as you might expect.
Interactive 2 — Hunting for the Pair $m,n$

To split the middle term we need two numbers with product $ac$ and sum $b$. The table on the left lists the factor pairs of $|ac|$ with the right signs, and the hunter checks their sums one by one: ✗ wrong sum, ✓ found. Then the grouping steps and the roots appear on the right. Press Play to run through several equations, or pick one and press Next pair ▸ to search by hand.

need—
pair being checked—
roots—

Word problems solved by factorisation

Now we can finish the modelling of Section 4.1: build the equation, factorise, and then check each root against the situation.

✏️ Example 5 — A right-angled triangle
The hypotenuse of a right triangle is $17$ cm. One of the other two sides is $7$ cm longer than the other. Find the lengths of these two sides.
Model Let the shorter side be $x$ cm; the other is $(x+7)$ cm. By Pythagoras, $x^2+(x+7)^2=17^2$, so $2x^2+14x+49=289$, i.e. $2x^2+14x-240=0$, or $x^2+7x-120=0$.
Factorise Sum $7$, product $-120$: $15$ and $-8$. $\;x^2+15x-8x-120=x(x+15)-8(x+15)=(x-8)(x+15)$.
Roots $x=8$ or $x=-15$. A length cannot be negative, so $x=8$.
Answer The sides are $8$ cm and $15$ cm. Check: $64+225=289=17^2$ ✓.
✏️ Example 6 — Buying pens
A shopkeeper spends ₹ $600$ on pens, all at the same price. If each pen had cost ₹ $5$ less, she could have bought $10$ more pens for the same money. Find the price of a pen and the number of pens she bought.
Model Let the price be ₹ $x$ per pen. Number of pens $=\dfrac{600}{x}$; at the lower price, $\dfrac{600}{x-5}$. So $\dfrac{600}{x-5}-\dfrac{600}{x}=10$.
Standard form Multiply by $x(x-5)$: $600x-600(x-5)=10x(x-5)$, so $3000=10x^2-50x$, i.e. $x^2-5x-300=0$.
Factorise Sum $-5$, product $-300$: $-20$ and $15$. $\;(x-20)(x+15)=0$, so $x=20$ or $x=-15$. A price is positive: $x=20$.
Answer Each pen costs ₹ $20$ and she bought $\tfrac{600}{20}=30$ pens. Check: at ₹ $15$ she would get $40$ pens — $10$ more ✓.

Exercises

I
Solve by factorisation.
  • 1 $x^2-3x-28=0$
  • 2 $x^2+11x+30=0$
  • 3 $2x^2-7x+3=0$
  • 4 $3x^2+5x-2=0$
  • 5 $6x^2+7x-3=0$
  • 6 $4x^2-12x+9=0$
  • 7 $25x^2-16=0$
  • 8 $5x^2=15x$
  • 9 $10x^2-x-21=0$
  • 10 $12x^2-25x+12=0$
Check: type whole-number coefficients and compare the pair, the factors and the roots with yours
pair $m,n$—
factorised—
roots—
1 Sum $-3$, product $-28$: $-7$, $4$. $(x-7)(x+4)=0$, so $x=7$ or $x=-4$.
2 Sum $11$, product $30$: $5$, $6$. $(x+5)(x+6)=0$, so $x=-5$ or $x=-6$.
3 $ac=6$, sum $-7$: $-6$, $-1$. $2x^2-6x-x+3=2x(x-3)-1(x-3)=(2x-1)(x-3)$. $x=\tfrac12$ or $x=3$.
4 $ac=-6$, sum $5$: $6$, $-1$. $3x^2+6x-x-2=3x(x+2)-1(x+2)=(3x-1)(x+2)$. $x=\tfrac13$ or $x=-2$.
5 $ac=-18$, sum $7$: $9$, $-2$. $6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)$. $x=\tfrac13$ or $x=-\tfrac32$.
6 $ac=36$, sum $-12$: $-6$, $-6$. $(2x-3)^2=0$, so $x=\tfrac32$ (a repeated root).
7 Difference of squares: $(5x-4)(5x+4)=0$, so $x=\tfrac45$ or $x=-\tfrac45$. (Splitting also works: $ac=-400$, sum $0$: $20$, $-20$.)
8 $5x^2-15x=0$, $\;5x(x-3)=0$, so $x=0$ or $x=3$. (Dividing by $x$ would lose the root $0$.)
9 $ac=-210$, sum $-1$: $-15$, $14$. $10x^2-15x+14x-21=5x(2x-3)+7(2x-3)=(5x+7)(2x-3)$. $x=\tfrac32$ or $x=-\tfrac75$.
10 $ac=144$, sum $-25$: $-16$, $-9$. $12x^2-16x-9x+12=4x(3x-4)-3(3x-4)=(4x-3)(3x-4)$. $x=\tfrac34$ or $x=\tfrac43$.
II
Rearrange first, then solve by factorisation.
  • 1 $x(x+4)=12$
  • 2 $(x-2)(x+3)=14$
  • 3 $(2x+1)^2=3x+4$
  • 4 $\sqrt2\,x^2-3x-2\sqrt2=0$
  • 5 $3x^2-2\sqrt6\,x+2=0$
  • 6 $x-\dfrac6x=1$  ($x\ne0$)
  • 7 $2x+\dfrac5x=11$  ($x\ne0$)
  • 8 $\dfrac4x-3=\dfrac{5}{2x+3}$  ($x\ne0$, $x\ne-\tfrac32$)
1 $x^2+4x-12=0$, $\;(x+6)(x-2)=0$: $x=-6$ or $x=2$.
2 $x^2+x-6=14$, so $x^2+x-20=0$, $\;(x+5)(x-4)=0$: $x=-5$ or $x=4$.
3 $4x^2+4x+1=3x+4$, so $4x^2+x-3=0$. $ac=-12$, sum $1$: $4$, $-3$. $4x(x+1)-3(x+1)=(4x-3)(x+1)$: $x=\tfrac34$ or $x=-1$.
4 $ac=\sqrt2\times(-2\sqrt2)=-4$, sum $-3$: $-4$, $1$. $\sqrt2\,x^2-4x+x-2\sqrt2=\sqrt2\,x(x-2\sqrt2)+1(x-2\sqrt2)=(\sqrt2\,x+1)(x-2\sqrt2)$ (using $4=\sqrt2\cdot2\sqrt2$). $x=2\sqrt2$ or $x=-\tfrac1{\sqrt2}=-\tfrac{\sqrt2}{2}$.
5 $ac=6$, sum $-2\sqrt6$: $-\sqrt6$, $-\sqrt6$. $3x^2-\sqrt6\,x-\sqrt6\,x+2=\sqrt3\,x(\sqrt3\,x-\sqrt2)-\sqrt2(\sqrt3\,x-\sqrt2)=(\sqrt3\,x-\sqrt2)^2$. Repeated root $x=\tfrac{\sqrt2}{\sqrt3}=\tfrac{\sqrt6}{3}$.
6 Multiply by $x$: $x^2-6=x$, so $x^2-x-6=0$, $\;(x-3)(x+2)=0$: $x=3$ or $x=-2$ (both non-zero, so both allowed).
7 Multiply by $x$: $2x^2+5=11x$, so $2x^2-11x+5=0$, $\;(2x-1)(x-5)=0$: $x=\tfrac12$ or $x=5$.
8 Multiply by $x(2x+3)$: $4(2x+3)-3x(2x+3)=5x$, so $8x+12-6x^2-9x-5x=0$, i.e. $-6x^2-6x+12=0$, or $x^2+x-2=0$. $(x+2)(x-1)=0$: $x=-2$ or $x=1$. Check $x=1$: $4-3=1$ and $\tfrac55=1$ ✓; $x=-2$: $-2-3=-5$ and $\tfrac{5}{-1}=-5$ ✓.
III
Word problems. Form a quadratic equation, solve it by factorisation and reject any root that does not fit.
  • 1 Two numbers differ by $3$ and their product is $180$. Find them.
  • 2 The sum of the squares of two consecutive natural numbers is $313$. Find the numbers.
  • 3 A rectangular hall is $5$ m longer than it is wide, and its floor area is $84$ m². Find its length and breadth.
  • 4 The hypotenuse of a right triangle is $10$ cm, and the other two sides differ by $2$ cm. Find them.
  • 5 A father is $30$ years older than his son, and the product of their present ages is $400$. Find their ages.
  • 6 A student spends ₹ $360$ on notebooks. Had each notebook been ₹ $2$ cheaper, she could have bought $6$ more. Find the price of a notebook.
  • 7 A boat goes $30$ km upstream and $30$ km back in $4$ hours $30$ minutes. Its speed in still water is $15$ km/h. Find the speed of the stream.
  • 8 In the park of Interactive 2 of Section 4.1, the grass satisfies $(20-2x)(15-2x)=126$. Solve by factorisation and find the width of the border.
  • 9 A ball is thrown straight up. Its height after $t$ seconds is $h=25t-5t^2$ metres. At what times is it $30$ m above the ground? Explain why there are two answers.
1 $x(x+3)=180$, $\;x^2+3x-180=0$, $\;(x+15)(x-12)=0$. The numbers are $12$ and $15$, or $-15$ and $-12$ (both pairs work: $(-15)(-12)=180$).
2 $x^2+(x+1)^2=313$, $\;2x^2+2x-312=0$, $\;x^2+x-156=0$, $\;(x+13)(x-12)=0$. Natural numbers: $12$ and $13$ ($144+169=313$ ✓).
3 $x(x+5)=84$, $\;x^2+5x-84=0$, $\;(x+12)(x-7)=0$. Breadth $7$ m, length $12$ m.
4 $x^2+(x+2)^2=100$, $\;2x^2+4x-96=0$, $\;x^2+2x-48=0$, $\;(x+8)(x-6)=0$. Sides $6$ cm and $8$ cm ($36+64=100$ ✓).
5 Son $x$ years: $x(x+30)=400$, $\;x^2+30x-400=0$, $\;(x+40)(x-10)=0$. Son $10$ years, father $40$ years.
6 Price ₹ $x$: $\dfrac{360}{x-2}-\dfrac{360}{x}=6$. Multiply by $x(x-2)$: $720=6x(x-2)$, $\;x^2-2x-120=0$, $\;(x-12)(x+10)=0$. Price ₹ $12$ ($30$ notebooks; at ₹ $10$, $36$ notebooks — $6$ more ✓).
7 Stream $y$ km/h: $\dfrac{30}{15-y}+\dfrac{30}{15+y}=\dfrac92$. Multiply by $2(225-y^2)$: $60(15+y)+60(15-y)=9(225-y^2)$, so $1800=2025-9y^2$, $\;9y^2-225=0$, $\;y^2-25=0$, $\;(y-5)(y+5)=0$. Stream $5$ km/h (upstream $3$ h at $10$ km/h, downstream $1.5$ h at $20$ km/h ✓).
8 $4x^2-70x+300=126$, so $2x^2-35x+87=0$. $ac=174$, sum $-35$: $-29$, $-6$. $2x^2-29x-6x+87=x(2x-29)-3(2x-29)=(x-3)(2x-29)$. $x=3$ or $x=14.5$; $14.5$ m is more than half of $15$ m, so the border is $3$ m wide.
9 $25t-5t^2=30$, $\;5t^2-25t+30=0$, $\;t^2-5t+6=0$, $\;(t-2)(t-3)=0$. At $t=2$ s (going up) and $t=3$ s (coming down).
IV
Harder problems.
  • 1 Find two consecutive positive odd integers whose product is $255$.
  • 2 One root of $x^2-kx+15=0$ is $3$. Find $k$, then factorise to find the other root.
  • 3 Solve for $x$: $\;x^2+2ax+a^2-b^2=0$.
  • 4 The product of the digits of a two-digit number is $24$. When $18$ is added to the number, its digits are reversed. Find the number.
1 $x(x+2)=255$, $\;x^2+2x-255=0$, $\;(x+17)(x-15)=0$. Positive: $15$ and $17$.
2 $9-3k+15=0$, so $k=8$. $x^2-8x+15=(x-3)(x-5)$: the other root is $5$.
3 $x^2+2ax+a^2=(x+a)^2$, so the equation is $(x+a)^2-b^2=0$, i.e. $(x+a-b)(x+a+b)=0$. Hence $x=b-a$ or $x=-a-b$. (Splitting the middle term with $m=a-b$, $n=a+b$ gives the same.)
4 Tens digit $t$, units digit $u$. Adding $18$ reverses the digits: $(10u+t)-(10t+u)=9(u-t)=18$, so $u=t+2$. Then $t(t+2)=24$, $\;t^2+2t-24=0$, $\;(t+6)(t-4)=0$, so $t=4$, $u=6$. The number is $46$ ($4\times6=24$ and $46+18=64$ ✓).
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE Main

How this is examined: board papers ask you to "find the roots by factorisation" (show the split middle term — it carries a mark) and set 3-mark word problems that end in a factorisable quadratic. JEE Main uses factorisation inside quick questions, often with a twist such as $|x|$. Always state which root you reject and why.

E1. Board · 1 mark The roots of $2x^2-x-6=0$ are  (A) $2,\ -\tfrac32$  (B) $-2,\ \tfrac32$  (C) $3,\ -2$  (D) $2,\ \tfrac32$

Factorise $ac=-12$, sum $-1$: $-4$, $3$. $2x^2-4x+3x-6=2x(x-2)+3(x-2)=(2x+3)(x-2)$, so $x=2$ or $x=-\tfrac32$.
Trap (B) is what you get if you read the roots straight off the brackets without changing sign: the factor $x-2$ gives the root $+2$, not $-2$.
Answer (A). Check: sum $\tfrac12=-\tfrac{-1}{2}$ ✓, product $-3=\tfrac{-6}{2}$ ✓.

E2. Board · 1 mark Assertion (A): The roots of $x^2=7x$ are $0$ and $7$. Reason (R): Dividing both sides of $x^2=7x$ by $x$ is always allowed, and gives $x=7$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check A $x^2-7x=0$, $\;x(x-7)=0$: $x=0$ or $x=7$. True.
Check R Dividing by $x$ is not allowed when $x=0$, and doing so loses the root $0$. False.
Answer (C).

E3. Board · 2 marks Solve for $x$: $\;4x^2-4ax+(a^2-b^2)=0$.

Split Product $4(a^2-b^2)=[2(a+b)][2(a-b)]$ and sum $-4a=-2(a+b)-2(a-b)$. So $4x^2-2(a+b)x-2(a-b)x+(a+b)(a-b)$. (1 mark)
Group $2x[2x-(a+b)]-(a-b)[2x-(a+b)]=[2x-(a+b)][2x-(a-b)]$.
Answer $x=\dfrac{a+b}{2}$ or $x=\dfrac{a-b}{2}$. (1 mark) (Quicker: $4x^2-4ax+a^2=(2x-a)^2$, so the equation is $(2x-a)^2-b^2=0$.)

E4. Board · 3 marks The sum of a number and its reciprocal is $\tfrac{10}{3}$. Find the number.

Model $x+\dfrac1x=\dfrac{10}{3}$ with $x\ne0$. Multiply by $3x$: $3x^2+3=10x$, so $3x^2-10x+3=0$. (1 mark)
Factorise $ac=9$, sum $-10$: $-9$, $-1$. $3x^2-9x-x+3=3x(x-3)-1(x-3)=(3x-1)(x-3)$. (1 mark)
Answer $x=3$ or $x=\tfrac13$ — the number is $3$ or $\tfrac13$. Check: $3+\tfrac13=\tfrac{10}{3}$ ✓. (1 mark)

E5. Board · 3 marks The diagonal of a rectangular tile is $8$ cm more than its shorter side, and its longer side is $7$ cm more than its shorter side. Find the sides of the tile.

Model Shorter side $x$ cm, longer side $(x+7)$ cm, diagonal $(x+8)$ cm. Pythagoras: $x^2+(x+7)^2=(x+8)^2$. (1 mark)
Simplify $2x^2+14x+49=x^2+16x+64$, so $x^2-2x-15=0$, $\;(x-5)(x+3)=0$. (1 mark)
Answer $x=-3$ is rejected (a length). Sides $5$ cm and $12$ cm, diagonal $13$ cm; check $25+144=169$ ✓. (1 mark)

E6. JEE Main The number of real solutions of $x^2-5|x|+6=0$ is ______. (numerical answer)

Idea $x^2=|x|^2$ for every real $x$. Put $y=|x|\ge0$: $\;y^2-5y+6=0$, $\;(y-2)(y-3)=0$, so $|x|=2$ or $|x|=3$.
Count $|x|=2$ gives $x=\pm2$; $|x|=3$ gives $x=\pm3$. Each of the four satisfies the equation (e.g. $x=-3$: $9-15+6=0$ ✓).
Answer $4$. The tempting answer $2$ forgets that each value of $|x|$ comes from two values of $x$.
🔬 Research–Teaching NexusMini project · optional, about 40 minutes

Research question. How wide must a uniform border around a photo be so that the framed picture has twice the area of the photo — and for which photo sizes does the equation factorise neatly?

Why it matters. Picture framers, poster designers and architects who put a path around a lawn or a margin around a page all face this question. For a photo $l$ cm by $w$ cm with a border $x$ cm wide, the framed size is $(l+2x)$ by $(w+2x)$, and "twice the area" gives the quadratic equation $(l+2x)(w+2x)=2lw$, i.e. $4x^2+2(l+w)x-lw=0$.

What to do.

  1. Measure (in cm) at least five rectangular pictures at home or school: a passport photo, a postcard, a certificate, a poster, a phone screen.
  2. For each one write the equation $4x^2+2(l+w)x-lw=0$ with your measurements.
  3. Try to solve each by factorisation. For measurements in whole centimetres, check first whether a suitable pair $m,n$ exists; if not, find $x$ to one decimal place by trial in a spreadsheet.
  4. Investigate the sizes $6\times4$, $12\times8$ and $18\times12$. What do you notice about the border width? Can you explain it?
  5. Cut a paper border of the width you found for one picture and check that the framed area really is double.

Tools: Ruler or tape measure, paper and scissors, calculator or spreadsheet

Hand in: A table with $l$, $w$, the equation, its factorisation (where possible) and the border width; your explanation of the pattern in step 4; and a photo of your paper frame.

Going further. Change the target from "twice the area" to "three times the area". Which photo sizes now give whole-number borders? (Section 4.3 gives a formula that solves the equations that do not factorise.)

Teacher/parent note: for $6\times4$ the equation is $x^2+5x-6=0$, root $x=1$; for $12\times8$ it is $x^2+10x-24=0$, root $x=2$; for $18\times12$ it is $x^2+15x-54=0$, root $x=3$. In general an $l\times w=6k\times4k$ photo needs a border of exactly $k$: $(6k+2k)(4k+2k)=48k^2=2\times24k^2$.

🧪 Self-check quiz

0/5 correct

Q1. The roots of $(x+4)(2x-3)=0$ are

Q2. To factorise $3x^2+10x+8$ by splitting the middle term, the two numbers are

Q3. The roots of $x^2-9x=0$ are

Q4. The roots of $4x^2-4x+1=0$ are

Q5. The product of two consecutive positive integers is $132$. The integers are

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