NCERT Class 10 Exercise 4.2 Solutions: Factorisation

Exer­cise 4.2 of the NCERT Class 10 chap­ter Qua­dratic Equa­tions is about one skill: solv­ing a qua­dratic equa­tion by fac­tori­sa­tion. It has six ques­tions. Ques­tion 1 gives five equa­tions to solve directly. Ques­tion 2 goes back to the two sit­u­a­tions of Exam­ple 1 (the mar­bles and the toys) and solves them. Ques­tions 3 to 6 are word prob­lems: you turn the words into a qua­dratic equa­tion, solve it, and then decide which root makes sense. Every answer below has been checked by putting it back into the orig­i­nal equa­tion or con­di­tion.

Quick recap: the method#

To solve $ax^2 + bx + c = 0$ by split­ting the mid­dle term:

1. Write the equa­tion in stan­dard form, with $0$ on the right. Clear frac­tions or com­mon fac­tors first if that makes the num­bers smaller.
2. Find two num­bers whose prod­uct is $ac$ and whose sum is $b$.
3. Split $bx$ into two terms using those num­bers.
4. Group the four terms in two pairs and take a com­mon fac­tor out of each pair. The two brack­ets left over must be the same.
5. Take that com­mon bracket out, so the left side becomes a prod­uct of two fac­tors.
6. Set each fac­tor equal to zero and solve.
7. For a word prob­lem, reject any root that does not fit the sit­u­a­tion (a neg­a­tive length, a neg­a­tive count, and so on).

Why this works: if the prod­uct of two real num­bers is zero, at least one of them must be zero. So $(x - p)(x - q) = 0$ holds exactly when $x = p$ or $x = q$. Those val­ues are the roots of the equa­tion, and they are also the points where the graph of $y = ax^2 + bx + c$ meets the $x$-axis.

Ques­tion 1: Find the roots of the equa­tions by fac­tori­sa­tion#

Ques­tion 1 (i)#

Solve $x^2 - 3x - 10 = 0$.

Here $a = 1$, $b = -3$, $c = -10$, so $ac = -10$. We need two num­bers with prod­uct $-10$ and sum $-3$. The pair $-5$ and $2$ works: $(-5)(2) = -10$ and $-5 + 2 = -3$.

$$\begin{aligned} x^2 - 5x + 2x - 10 &= 0 \\ x(x - 5) + 2(x - 5) &= 0 \\ (x - 5)(x + 2) &= 0 \end{aligned}$$

So $x - 5 = 0$ or $x + 2 = 0$, which gives $x = 5$ or $x = -2$.

Check: $5^2 - 3(5) - 10 = 25 - 15 - 10 = 0$, and $(-2)^2 - 3(-2) - 10 = 4 + 6 - 10 = 0$.

Answer: The roots are $5$ and $-2$.

Ques­tion 1 (ii)#

Solve $2x^2 + x - 6 = 0$.

Here $a = 2$, $b = 1$, $c = -6$, so $ac = -12$. We need two num­bers with prod­uct $-12$ and sum $1$: these are $4$ and $-3$.

$$\begin{aligned} 2x^2 + 4x - 3x - 6 &= 0 \\ 2x(x + 2) - 3(x + 2) &= 0 \\ (x + 2)(2x - 3) &= 0 \end{aligned}$$

So $x + 2 = 0$ or $2x - 3 = 0$, which gives $x = -2$ or $\displaystyle x = \frac{3}{2}$.

Check: $2(-2)^2 + (-2) - 6 = 8 - 2 - 6 = 0$, and $\displaystyle 2\left(\frac{3}{2}\right)^2 + \frac{3}{2} - 6 = \frac{9}{2} + \frac{3}{2} - 6 = 6 - 6 = 0$.

Answer: The roots are $-2$ and $\displaystyle \frac{3}{2}$.

Ques­tion 1 (iii)#

Solve $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$.

The coef­fi­cients con­tain surds, but the rule is the same. Here $a = \sqrt{2}$ and $c = 5\sqrt{2}$, so

$$ac = \sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10.$$

We need two num­bers with prod­uct $10$ and sum $7$: these are $5$ and $2$.

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

Now group the terms. In the first pair take out $x$: $\sqrt{2}\,x^2 + 5x = x(\sqrt{2}\,x + 5)$. In the sec­ond pair take out $\sqrt{2}$, using $2 = \sqrt{2} \times \sqrt{2}$: $2x + 5\sqrt{2} = \sqrt{2}(\sqrt{2}\,x + 5)$. The brack­ets match, so

$$\begin{aligned} x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) &= 0 \\ (\sqrt{2}\,x + 5)(x + \sqrt{2}) &= 0 \end{aligned}$$

So $\sqrt{2}\,x + 5 = 0$ or $x + \sqrt{2} = 0$. The first gives $\displaystyle x = -\frac{5}{\sqrt{2}}$. Ratio­nal­is­ing the denom­i­na­tor, $\displaystyle -\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$. The sec­ond gives $x = -\sqrt{2}$.

Check: for $x = -\sqrt{2}$: $\sqrt{2}(2) + 7(-\sqrt{2}) + 5\sqrt{2} = 2\sqrt{2} - 7\sqrt{2} + 5\sqrt{2} = 0$. For $\displaystyle x = -\frac{5}{\sqrt{2}}$: $\displaystyle x^2 = \frac{25}{2}$, so $\displaystyle \sqrt{2} \times \frac{25}{2} + 7\left(-\frac{5}{\sqrt{2}}\right) + 5\sqrt{2} = \frac{25\sqrt{2}}{2} - \frac{35\sqrt{2}}{2} + \frac{10\sqrt{2}}{2} = 0$. (We used $\displaystyle \frac{35}{\sqrt{2}} = \frac{35\sqrt{2}}{2}$ and $\displaystyle 5\sqrt{2} = \frac{10\sqrt{2}}{2}$.)

Answer: The roots are $-\sqrt{2}$ and $\displaystyle -\frac{5}{\sqrt{2}}$, that is, $\displaystyle -\frac{5\sqrt{2}}{2}$.

Ques­tion 1 (iv)#

Solve $\displaystyle 2x^2 - x + \frac{1}{8} = 0$.

First clear the frac­tion by mul­ti­ply­ing every term by $8$. Mul­ti­ply­ing both sides by a non-zero num­ber does not change the roots.

$$16x^2 - 8x + 1 = 0$$

Now $ac = 16$ and $b = -8$, so we need two num­bers with prod­uct $16$ and sum $-8$: these are $-4$ and $-4$.

$$\begin{aligned} 16x^2 - 4x - 4x + 1 &= 0 \\ 4x(4x - 1) - 1(4x - 1) &= 0 \\ (4x - 1)(4x - 1) &= 0 \end{aligned}$$

Both fac­tors are the same, so $(4x - 1)^2 = 0$ and $4x - 1 = 0$, giv­ing $\displaystyle x = \frac{1}{4}$. The equa­tion has two equal roots, $\displaystyle \frac{1}{4}$ and $\displaystyle \frac{1}{4}$.

Check: $\displaystyle 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} + \frac{1}{8} = \frac{1}{8} - \frac{2}{8} + \frac{1}{8} = 0$.

Answer: The roots are $\displaystyle \frac{1}{4}$ and $\displaystyle \frac{1}{4}$ (equal roots).

Ques­tion 1 (v)#

Solve $100x^2 - 20x + 1 = 0$.

Here $ac = 100$ and $b = -20$, so we need two num­bers with prod­uct $100$ and sum $-20$: these are $-10$ and $-10$.

$$\begin{aligned} 100x^2 - 10x - 10x + 1 &= 0 \\ 10x(10x - 1) - 1(10x - 1) &= 0 \\ (10x - 1)(10x - 1) &= 0 \end{aligned}$$

So $(10x - 1)^2 = 0$, which gives $\displaystyle x = \frac{1}{10}$ twice. You can also spot this directly: $100x^2 - 20x + 1 = (10x)^2 - 2(10x)(1) + 1^2 = (10x - 1)^2$.

Check: $\displaystyle 100\left(\frac{1}{10}\right)^2 - 20\left(\frac{1}{10}\right) + 1 = 1 - 2 + 1 = 0$.

Answer: The roots are $\displaystyle \frac{1}{10}$ and $\displaystyle \frac{1}{10}$ (equal roots).

Ques­tions 3 to 6: word prob­lems#

Ques­tion 3#

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

Let one num­ber be $x$. Then the other is $27 - x$, and

$$\begin{aligned} x(27 - x) &= 182 \\ 27x - x^2 - 182 &= 0 \\ x^2 - 27x + 182 &= 0 \end{aligned}$$

We need two num­bers with prod­uct $182$ and sum $-27$. Since $182 = 2 \times 7 \times 13 = 13 \times 14$ and $13 + 14 = 27$, the pair is $-13$ and $-14$.

$$\begin{aligned} x^2 - 13x - 14x + 182 &= 0 \\ x(x - 13) - 14(x - 13) &= 0 \\ (x - 13)(x - 14) &= 0 \end{aligned}$$

So $x = 13$ or $x = 14$. If $x = 13$, the other num­ber is $27 - 13 = 14$; if $x = 14$, the other is $13$. Either way the pair is the same.

Check: $13 + 14 = 27$ and $13 \times 14 = 182$.

Answer: The two num­bers are $13$ and $14$.

Ques­tion 4#

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

Let the smaller inte­ger be $x$. The next one is $x + 1$, so

$$\begin{aligned} x^2 + (x + 1)^2 &= 365 \\ x^2 + x^2 + 2x + 1 &= 365 \\ 2x^2 + 2x - 364 &= 0 \\ x^2 + x - 182 &= 0 \end{aligned}$$

(In the last step we divided every term by $2$.) We need two num­bers with prod­uct $-182$ and sum $1$: these are $14$ and $-13$.

$$\begin{aligned} x^2 + 14x - 13x - 182 &= 0 \\ x(x + 14) - 13(x + 14) &= 0 \\ (x + 14)(x - 13) &= 0 \end{aligned}$$

So $x = -14$ or $x = 13$. The ques­tion asks for pos­i­tive inte­gers, so we reject $x = -14$ and take $x = 13$. The next inte­ger is $14$.

Check: $13^2 + 14^2 = 169 + 196 = 365$.

Why this works: the rejected root is not wrong alge­bra. The pair $-14, -13$ also has squares adding to $365$, but those inte­gers are not pos­i­tive, so the con­di­tion in the ques­tion rules them out.

Answer: The inte­gers are $13$ and $14$.

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.

Let the base be $x$ cm. Then the alti­tude is $(x - 7)$ cm. The base and alti­tude are the two sides that form the right angle, so by Pythago­ras' the­o­rem

$$\begin{aligned} x^2 + (x - 7)^2 &= 13^2 \\ x^2 + x^2 - 14x + 49 &= 169 \\ 2x^2 - 14x - 120 &= 0 \\ x^2 - 7x - 60 &= 0 \end{aligned}$$

We need two num­bers with prod­uct $-60$ and sum $-7$: these are $-12$ and $5$.

$$\begin{aligned} x^2 - 12x + 5x - 60 &= 0 \\ x(x - 12) + 5(x - 12) &= 0 \\ (x - 12)(x + 5) &= 0 \end{aligned}$$

So $x = 12$ or $x = -5$. A length can­not be neg­a­tive, so $x = 12$. The alti­tude is $12 - 7 = 5$ cm.

Check: $12^2 + 5^2 = 144 + 25 = 169 = 13^2$, and $5$ is indeed $7$ less than $12$.

Answer: The base is $12$ cm and the alti­tude is $5$ cm.

Ques­tion 6#

A cot­tage indus­try makes a cer­tain num­ber of pot­tery arti­cles in a day. On one day, the cost of mak­ing each arti­cle (in rupees) was $3$ more than twice the num­ber of arti­cles made that day, and the total cost of pro­duc­tion was ₹90. Find the num­ber of arti­cles made and the cost of each arti­cle.

Let the num­ber of arti­cles be $x$. The cost of each arti­cle is ₹$(2x + 3)$, so the total cost is

$$\begin{aligned} x(2x + 3) &= 90 \\ 2x^2 + 3x - 90 &= 0 \end{aligned}$$

Here $ac = 2 \times (-90) = -180$ and $b = 3$. We need two num­bers with prod­uct $-180$ and sum $3$: these are $15$ and $-12$.

$$\begin{aligned} 2x^2 + 15x - 12x - 90 &= 0 \\ x(2x + 15) - 6(2x + 15) &= 0 \\ (2x + 15)(x - 6) &= 0 \end{aligned}$$

So $\displaystyle x = -\frac{15}{2}$ or $x = 6$. The num­ber of arti­cles must be a pos­i­tive whole num­ber, so we reject $\displaystyle -\frac{15}{2}$ and take $x = 6$. The cost of each arti­cle is $2(6) + 3 = 15$ rupees.

Check: $6 \times 15 = 90$, and $2(6)^2 + 3(6) - 90 = 72 + 18 - 90 = 0$.

Answer: $6$ arti­cles were made, and each cost ₹15.

Com­mon mis­takes#

• Look­ing for two num­bers whose prod­uct is $c$ when $a \ne 1$. The prod­uct must be $ac$; in Ques­tion 1 (ii) it is $-12$, not $-6$.
• Get­ting a sign wrong when tak­ing out a neg­a­tive com­mon fac­tor: $-3x - 6 = -3(x + 2)$, not $-3(x - 2)$.
• Divid­ing both sides by $x$ to "sim­plify". That can throw away a root; always bring every­thing to one side and fac­torise.
• Writ­ing only one root when the two fac­tors are equal. Say the equa­tion has two equal roots, as in Ques­tions 1 (iv) and 1 (v).
• For­get­ting to reject a root that does not fit the prob­lem, such as a neg­a­tive length in Ques­tion 5 or a frac­tional num­ber of arti­cles in Ques­tion 6.
• Stop­ping at $x$ in a word prob­lem. Ques­tion 6 also asks for the cost of each arti­cle, and Ques­tion 5 asks for both sides.

Key terms#

Qua­dratic equa­tion

An equa­tion of the form $ax^2 + bx + c = 0$ where $a$, $b$, $c$ are real num­bers and $a \ne 0$.

Root

A value of $x$ that makes $ax^2 + bx + c$ equal to zero; a qua­dratic equa­tion has at most two roots.

Fac­tori­sa­tion

Writ­ing an expres­sion as a prod­uct of sim­pler expres­sions, such as $x^2 - 3x - 10 = (x - 5)(x + 2)$.

Split­ting the mid­dle term

Replac­ing $bx$ by two terms whose coef­fi­cients mul­ti­ply to $ac$ and add to $b$, so the expres­sion can be grouped and fac­torised.

Zero-prod­uct prop­erty

If $pq = 0$ for real num­bers $p$ and $q$, then $p = 0$ or $q = 0$.

Equal roots

Two roots with the same value, which hap­pens when the qua­dratic is a per­fect square, such as $(4x - 1)^2$.

Stan­dard form

The arrange­ment $ax^2 + bx + c = 0$ with all terms on one side, in descend­ing pow­ers of $x$.

Ratio­nal­is­ing the denom­i­na­tor

Remov­ing a surd from a denom­i­na­tor, for exam­ple $\displaystyle \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$.

Com­mon ques­tions#

Why must the two num­bers mul­ti­ply to $ac$ and not to $c$?#

When you group $ax^2 + px + qx + c$, the first pair has com­mon fac­tor involv­ing $a$ and the sec­ond involves $c$. The brack­ets only match when $pq = ac$. When $a = 1$, $ac$ is sim­ply $c$, which is why the short­cut seems to work for equa­tions like $x^2 - 3x - 10 = 0$.

How do I find the pair of num­bers quickly?#

List the fac­tor pairs of $|ac|$. If $ac$ is pos­i­tive, both num­bers have the same sign as $b$; if $ac$ is neg­a­tive, the num­bers have oppo­site signs and the larger one takes the sign of $b$.

Can a qua­dratic have only one root?#

It can have two equal roots, as in $16x^2 - 8x + 1 = 0$, where both roots are $\displaystyle \frac{1}{4}$. In NCERT lan­guage we still say it has two roots, and they are equal.

Is it wrong to mul­ti­ply or divide the whole equa­tion by a num­ber?#

No. Mul­ti­ply­ing or divid­ing both sides by the same non-zero num­ber leaves the roots unchanged. That is how we cleared the frac­tion in Ques­tion 1 (iv) and halved the equa­tions in Ques­tions 4 and 5.

What if I can­not find two suit­able num­bers?#

Then the qua­dratic may not fac­torise neatly over the inte­gers, or it may have no real roots at all. The qua­dratic for­mula and the dis­crim­i­nant, cov­ered in the next exer­cise, han­dle every case.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.