NCERT Class 10 Mathematics — Quadratic Equations, Exercise 4.2. All 6 questions solved.

Exercise 4.2 is about one method, factorisation, used first on bare equations and then on word problems. Everything rests on two facts:

• Splitting the middle term. To factorise $ax^2 + bx + c$, find two numbers whose product is $ac$ and whose sum is $b$, and use them to split $bx$ into two terms that can be grouped in pairs.
• The zero-product rule. If $PQ = 0$ then $P = 0$ or $Q = 0$. That is what turns a factorised quadratic into two separate linear equations.

In the word problems there is a third step that the algebra cannot do for you: deciding which roots actually make sense as a length, an age, or a count.

Key insight. Factorising is only useful because the right-hand side is $0$. A product equal to $0$ forces one factor to be $0$; a product equal to, say, $124$ tells you nothing about either factor. So every question here starts by moving everything to one side, and only then factorises.

Question 1

Find the roots of the following quadratic equations by factorisation:

(i) $x^2 – 3x – 10 = 0$    (ii) $2x^2 + x – 6 = 0$    (iii) $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$

(iv) $2x^2 – x + \dfrac{1}{8} = 0$    (v) $100x^2 – 20x + 1 = 0$

Solution. In each part, find the pair with product $ac$ and sum $b$, split the middle term, and group.

(i) Here $ac = -10$ and $b = -3$. The pair is $-5$ and $2$, since $(-5)(2) = -10$ and $-5 + 2 = -3$:

$$x^2 – 5x + 2x – 10 = x(x – 5) + 2(x – 5) = (x – 5)(x + 2) = 0$$

So $x – 5 = 0$ or $x + 2 = 0$, giving $x = 5$ or $x = -2$.

(ii) Now $ac = 2 \times (-6) = -12$ and $b = 1$. The pair is $4$ and $-3$:

$$2x^2 + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2) = (2x – 3)(x + 2) = 0$$

So $x = \dfrac{3}{2}$ or $x = -2$.

(iii) The surds look awkward, but the rule is unchanged. The product is $ac = \sqrt{2} \times 5\sqrt{2} = 10$, and the sum must be $7$, so the pair is $5$ and $2$. Split $7x$ as $5x + 2x$ and group so that each pair shares a factor:

$$\sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5)$$

The second group works because $\sqrt{2} \times \sqrt{2}\,x = 2x$. So

$$(\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0$$

giving $x = -\dfrac{5}{\sqrt{2}} = -\dfrac{5\sqrt{2}}{2}$ or $x = -\sqrt{2}$.

(iv) A fraction among the coefficients is easiest removed first. Multiplying the whole equation by $8$ does not change its roots:

$$16x^2 – 8x + 1 = 0$$

Now $ac = 16$ and $b = -8$, so the pair is $-4$ and $-4$:

$$16x^2 – 4x – 4x + 1 = 4x(4x – 1) – 1(4x – 1) = (4x – 1)^2 = 0$$

Both factors are the same, so both give $x = \dfrac{1}{4}$. The equation has the repeated root $\dfrac{1}{4}, \dfrac{1}{4}$.

(v) Here $ac = 100$ and $b = -20$, so the pair is $-10$ and $-10$:

$$100x^2 – 10x – 10x + 1 = 10x(10x – 1) – 1(10x – 1) = (10x – 1)^2 = 0$$

Again a perfect square, so the root $\dfrac{1}{10}$ is repeated.

Question 2

Solve the problems given in Example 1.

Example 1 of the chapter set up two situations as quadratic equations:

(i) John and Jivanti together have $45$ marbles. Both of them lost $5$ marbles each, and the product of the number of marbles they now have is $124$. How many marbles did they have to start with?

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $55$ minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ $750$. How many toys were produced on that day?

Solution.

(i) If John had $x$ marbles, Jivanti had $45 – x$. After losing $5$ each they have $x – 5$ and $40 – x$, and the example showed that $(x – 5)(40 – x) = 124$ reduces to

$$x^2 – 45x + 324 = 0$$

We need two numbers with product $324$ and sum $-45$: they are $-9$ and $-36$.

$$x^2 – 9x – 36x + 324 = x(x – 9) – 36(x – 9) = (x – 9)(x – 36) = 0$$

So $x = 9$ or $x = 36$. Both are acceptable. If John had $9$, Jivanti had $45 – 9 = 36$; if John had $36$, Jivanti had $9$. Either way, after the losses they hold $4$ and $31$, and $4 \times 31 = 124$ as required.

(ii) With $x$ toys, each costs ₹ $(55 – x)$, and the example reduced $x(55 – x) = 750$ to

$$x^2 – 55x + 750 = 0$$

The pair with product $750$ and sum $-55$ is $-25$ and $-30$:

$$x^2 – 25x – 30x + 750 = x(x – 25) – 30(x – 25) = (x – 25)(x – 30) = 0$$

So $x = 25$ or $x = 30$. Both work: $25$ toys at ₹ $30$ each cost ₹ $750$, and so do $30$ toys at ₹ $25$ each.

Question 3

Find two numbers whose sum is $27$ and product is $182$.

Solution. Two unknowns with a known sum can be written with one letter: call the numbers $x$ and $27 – x$. Their product is $182$:

$$x(27 – x) = 182 \quad\Longrightarrow\quad x^2 – 27x + 182 = 0$$

The pair with product $182$ and sum $-27$ is $-13$ and $-14$:

$$x^2 – 13x – 14x + 182 = x(x – 13) – 14(x – 13) = (x – 13)(x – 14) = 0$$

So $x = 13$ or $x = 14$. The two roots describe the same pair of numbers: if $x = 13$ the other is $14$, and if $x = 14$ the other is $13$.

Question 4

Find two consecutive positive integers, sum of whose squares is $365$.

Solution. Consecutive integers differ by $1$, so call them $x$ and $x + 1$:

$$x^2 + (x + 1)^2 = 365 \quad\Longrightarrow\quad 2x^2 + 2x – 364 = 0 \quad\Longrightarrow\quad x^2 + x – 182 = 0$$

Dividing by $2$ keeps the numbers small. The pair with product $-182$ and sum $1$ is $14$ and $-13$:

$$x^2 + 14x – 13x – 182 = x(x + 14) – 13(x + 14) = (x + 14)(x – 13) = 0$$

So $x = 13$ or $x = -14$. The question asks for positive integers, so $x = -14$ is rejected (it would give the pair $-14, -13$, which does satisfy the equation but not the question). With $x = 13$ the integers are $13$ and $14$, and indeed $169 + 196 = 365$.

Question 5

The altitude of a right triangle is $7$ cm less than its base. If the hypotenuse is $13$ cm, find the other two sides.

Solution. In a right triangle the base and the altitude are the two sides that meet at the right angle, so Pythagoras’ theorem links them to the hypotenuse. Let the base be $x$ cm; then the altitude is $(x – 7)$ cm and

$$x^2 + (x – 7)^2 = 13^2 \quad\Longrightarrow\quad 2x^2 – 14x + 49 = 169 \quad\Longrightarrow\quad x^2 – 7x – 60 = 0$$

The pair with product $-60$ and sum $-7$ is $-12$ and $5$:

$$x^2 – 12x + 5x – 60 = x(x – 12) + 5(x – 12) = (x – 12)(x + 5) = 0$$

So $x = 12$ or $x = -5$. A length cannot be negative, so $x = 12$: the base is $12$ cm and the altitude $12 – 7 = 5$ cm. Check: $12^2 + 5^2 = 144 + 25 = 169 = 13^2$.

Question 6

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ $90$, find the number of articles produced and the cost of each article.

Solution. Let $x$ articles be produced. “Three more than twice the number” makes the cost of each article ₹ $(2x + 3)$, and total cost is number times cost each:

$$x(2x + 3) = 90 \quad\Longrightarrow\quad 2x^2 + 3x – 90 = 0$$

Here $ac = -180$ and $b = 3$, so the pair is $15$ and $-12$:

$$2x^2 + 15x – 12x – 90 = x(2x + 15) – 6(2x + 15) = (2x + 15)(x – 6) = 0$$

So $x = 6$ or $x = -\dfrac{15}{2}$. A number of articles must be a positive whole number, so $x = 6$, and each article costs ₹ $(2 \times 6 + 3) =$ ₹ $15$. Check: $6 \times 15 = 90$.

Common mistakes

• Question 1(iii), taking the product as $5$ instead of $10$. The product to split on is $a \times c = \sqrt{2} \times 5\sqrt{2}$, and the two surds multiply to $2$. Using only the rational parts gives $5$, and no pair of numbers with product $5$ adds to $7$ — which is usually where students decide the equation cannot be factorised.
• Question 1(ii), reading the roots straight off the factors with the wrong sign. The factor $(2x – 3)$ gives $x = \frac{3}{2}$, not $-\frac{3}{2}$ and not $3$. A root is the value that makes the factor zero, so solve each factor as its own small equation.
• Question 1(iv), factorising with the fraction still in place. Splitting $-x$ into two terms with product $\frac{2}{8}$ is possible but error-prone. Multiplying through by $8$ first changes nothing about the roots and leaves whole numbers to work with.
• Question 1(iv) and (v), writing a single root. A perfect square $(4x – 1)^2 = 0$ still comes from a quadratic, which has two roots; here they happen to coincide. The book lists them as $\frac{1}{4}, \frac{1}{4}$, and the next exercise explains why this happens.
• Questions 4, 5 and 6, keeping the negative root. Algebra solves the equation, not the problem. $-14$ is not a positive integer, $-5$ cm is not a side, and $-7.5$ articles is not a count; each needs to be rejected with a reason, not silently dropped.
• Question 2(i), answering “9 and 36” without saying whose is whose. Both roots are valid here, and they are the two ways of assigning the same pair of totals to John and Jivanti. Saying so is part of the answer.

Practise next

• Exercise 4.3 — the discriminant, which explains in advance why question 1(iv) and (v) had a repeated root and predicts when factorisation will fail.
• Exercise 4.1 — revise how the word problems of question 2 there are turned into equations; solving them by factorisation is good practice for this exercise.