Applications and Word Problems

A Strategy for Word Problems

Word problems are where quadratic equations earn their keep. The reliable recipe:

1. Read carefully and identify the unknown. Let it be $x$ (with units).
2. Translate every condition into algebra; combine into one equation in $x$.
3. Rearrange to standard form $ax^2 + bx + c = 0$.
4. Solve (factorise / formula).
5. Reject inadmissible roots (negative length, fractional people, etc.) and state the answer in words.

Key Point: The final, crucial step is checking which root makes sense. A quadratic gives two roots, but the situation usually allows only one.

[Board Important] Stating 'Let … be $x$' and rejecting the impossible root both carry marks. Never skip the interpretation step.

Number and Age Problems

These translate directly into quadratics.

Number type

'The sum of the squares of two consecutive natural numbers is 313.' Let them be $x$ and $x+1$: $x^2 + (x+1)^2 = 313 \Rightarrow 2x^2 + 2x - 312 = 0 \Rightarrow x^2 + x - 156 = 0$.

Age type

'A girl's present age squared, minus her age 5 years ago times 5, equals …' — express each age relative to the present ($x$, $x \pm n$) and form the equation.

Key Point: 'Consecutive' integers are $x, x+1$; 'consecutive even/odd' are $x, x+2$. Squares and products of these give quadratics.

[Board Important] For natural-number answers, reject negative or non-integer roots — only the sensible value is the final answer.

Speed, Distance and Time Problems

These are the most common quadratic word problems in board exams. The key relation is:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Many problems say a speed changed, or a journey took a different time. Setting up the time difference gives a quadratic.

Worked outline

'A train travels 360 km at uniform speed. If the speed were 5 km/h more, it would take 1 hour less.'

• Let speed $= x$ km/h. Time at speed $x$: $\dfrac{360}{x}$; at speed $x+5$: $\dfrac{360}{x+5}$.
• Condition: $\dfrac{360}{x} - \dfrac{360}{x+5} = 1$.
• This simplifies to $x^2 + 5x - 1800 = 0$, giving $x = 40$ (rejecting $-45$).

Key Point: Write each time as distance/speed, set up the given time difference, and clear denominators to reach a quadratic.

[Board Important] After solving, reject negative speeds. State the answer with units (km/h).

Area, Geometry and Work Problems

Area / geometry

Use area = length × breadth, or the Pythagoras theorem, to form the equation. 'The hypotenuse of a right triangle is 13 cm and one side is 7 cm more than the other.' Let the shorter side $= x$: $x^2 + (x+7)^2 = 13^2$.

Time-and-work / pipes

If one tap fills a tank in $x$ hours and another in $x + a$ hours, their combined rate gives a quadratic when a total time is specified.

Key Point: Identify the right formula (area, Pythagoras, or rate = 1/time), substitute the relationship between the unknowns, and simplify to standard form.

[Board Important] Pythagoras-based problems are very common: $a^2 + b^2 = c^2$, with one side expressed in terms of the other, leads straight to a quadratic.

Solved Examples

Example 1: Consecutive integers

The product of two consecutive positive integers is 306. Find them.

Solution:

1. Let them be $x$ and $x+1$: $x(x+1) = 306 \Rightarrow x^2 + x - 306 = 0$.
2. $(x + 18)(x - 17) = 0 \Rightarrow x = 17$ (reject $-18$).
3. The integers are 17 and 18.

Final Answer: 17 and 18.

Takeaway: Reject the negative root for positive integers.

Example 2: Sum of squares

The sum of squares of two consecutive natural numbers is 313. Find them.

Solution:

1. $x^2 + (x+1)^2 = 313 \Rightarrow 2x^2 + 2x + 1 = 313 \Rightarrow x^2 + x - 156 = 0$.
2. $(x + 13)(x - 12) = 0 \Rightarrow x = 12$.
3. The numbers are 12 and 13.

Final Answer: 12 and 13.

Takeaway: Expand $(x+1)^2$ and simplify to standard form.

Example 3: Rectangle area

The length of a rectangle is 5 m more than its breadth, and its area is 84 m². Find the dimensions.

Solution:

1. Let breadth $= x$: $x(x + 5) = 84 \Rightarrow x^2 + 5x - 84 = 0$.
2. $(x + 12)(x - 7) = 0 \Rightarrow x = 7$ (reject $-12$).
3. Breadth 7 m, length 12 m.

Final Answer: Breadth 7 m, length 12 m.

Takeaway: Area gives the product; reject the negative root.

Example 4: Train speed

A train travels 360 km at uniform speed. If the speed were 5 km/h more, it would take 1 hour less. Find the speed.

Solution:

1. Let speed $= x$. $\dfrac{360}{x} - \dfrac{360}{x + 5} = 1$.
2. $360(x + 5) - 360x = x(x+5) \Rightarrow 1800 = x^2 + 5x$.
3. $x^2 + 5x - 1800 = 0 \Rightarrow (x + 45)(x - 40) = 0 \Rightarrow x = 40$.

Final Answer: Speed $= 40$ km/h.

Takeaway: Time difference = 1 hour gives the quadratic; reject negative speed.

Example 5: Pythagoras setup

The hypotenuse of a right triangle is 13 cm. One leg is 7 cm longer than the other. Find the legs.

Solution:

1. Let the shorter leg $= x$: $x^2 + (x + 7)^2 = 13^2$.
2. $x^2 + x^2 + 14x + 49 = 169 \Rightarrow 2x^2 + 14x - 120 = 0 \Rightarrow x^2 + 7x - 60 = 0$.
3. $(x + 12)(x - 5) = 0 \Rightarrow x = 5$.
4. Legs are 5 cm and 12 cm.

Final Answer: 5 cm and 12 cm.

Takeaway: Pythagoras gives $a^2 + b^2 = c^2$; substitute the relation between legs.

Example 6: Age problem

The product of a girl's age 5 years ago and her age 8 years later is 30. Find her present age.

Solution:

1. Let present age $= x$: $(x - 5)(x + 8) = 30$.
2. $x^2 + 3x - 40 = 30 \Rightarrow x^2 + 3x - 70 = 0$.
3. $(x + 10)(x - 7) = 0 \Rightarrow x = 7$ (reject $-10$).

Final Answer: Present age 7 years.

Takeaway: Express both ages relative to the present, multiply, and reject the negative root.

Example 7: Two pipes / taps

Two taps together fill a tank in 6 hours. The larger tap takes 5 hours less than the smaller. Find the time each takes alone.

Solution:

1. Let the smaller tap take $x$ hours; larger takes $x - 5$. Combined rate: $\dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{1}{6}$.
2. $6(x - 5) + 6x = x(x - 5) \Rightarrow 12x - 30 = x^2 - 5x \Rightarrow x^2 - 17x + 30 = 0$.
3. $(x - 15)(x - 2) = 0 \Rightarrow x = 15$ (reject $x = 2$, since $x - 5$ would be negative).

Final Answer: Smaller tap 15 hours, larger tap 10 hours.

Takeaway: Reject roots that make a time negative.

Example 8: Marks problem

The sum of the squares of a student's marks in two subjects is 169, and the marks differ by 7. Find the marks.

Solution:

1. Let the larger mark be $x$ and the smaller $x - 7$: $x^2 + (x - 7)^2 = 169$.
2. $2x^2 - 14x + 49 = 169 \Rightarrow 2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0$.
3. $(x - 12)(x + 5) = 0 \Rightarrow x = 12$ (reject $-5$).
4. The marks are 12 and $12 - 7 = 5$.

Final Answer: The marks are 12 and 5.

Takeaway: Let the two quantities be $x$ and $x - 7$; reject the negative root.

Example 9: Reciprocal/number problem

The sum of a number and twice its reciprocal is 3. Find the number.

Solution:

1. Let the number be $x$: $x + \dfrac{2}{x} = 3$.
2. Multiply by $x$: $x^2 + 2 = 3x \Rightarrow x^2 - 3x + 2 = 0$.
3. $(x - 1)(x - 2) = 0 \Rightarrow x = 1$ or $x = 2$.

Final Answer: The number is 1 or 2.

Takeaway: Clear the denominator (multiply by $x$) to turn a reciprocal condition into a quadratic.

Example 10: Stream/boat speed

A motorboat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than to return downstream. Find the speed of the stream.

Solution:

1. Let stream speed $= x$. Upstream speed $= 18 - x$, downstream $= 18 + x$.
2. $\dfrac{24}{18 - x} - \dfrac{24}{18 + x} = 1$.
3. $24(18 + x) - 24(18 - x) = (18 - x)(18 + x) \Rightarrow 48x = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0$.
4. $(x + 54)(x - 6) = 0 \Rightarrow x = 6$ (reject $-54$).

Final Answer: Stream speed $= 6$ km/h.

Takeaway: Upstream/downstream speeds are $18 \mp x$; the time difference gives the quadratic.