NCERT Solutions for Class 9 Maths Chapter 4 — End-of-Chapter Exercises#

We write $-3x + 4$ as $4 - 3x$ and use $(a - b)^2$ with $a = 4$, $b = 3x$. We expand $(-3x + 4)^2$:

$$= 4^2 - 2(4)(3x) + (3x)^2$$

$$= 16 - 24x + 9x^2$$

$$= 9x^2 - 24x + 16$$

Using $(a + b)(a - b) = a^2 - b^2$ with $a = 2s$, $b = 7$. We find $(2s + 7)(2s - 7)$:

$$= (2s)^2 - 7^2$$

$$= 4s^2 - 49$$

Using $(a + b)(a - b) = a^2 - b^2$ with $a = p^2$, $b = \dfrac{1}{2}$. We find $\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)$:

$$= (p^2)^2 - \left(\dfrac{1}{2}\right)^2$$

$$= p^4 - \dfrac{1}{4}$$

Using $(a + b)(a - b) = a^2 - b^2$ with $a = 2n$, $b = 7$. We find $(2n + 7)(2n - 7)$:

$$= (2n)^2 - 7^2$$

$$= 4n^2 - 49$$

This is $(a - b)(a^2 + ab + b^2) = a^3 - b^3$ with $a = s$, $b = 2t$ (since $ab = 2st$ and $b^2 = 4t^2$). We find $(s - 2t)(s^2 + 2st + 4t^2)$:

$$= s^3 - (2t)^3$$

$$= s^3 - 8t^3$$

Using $(a - b)^2$ with $a = \dfrac{1}{2r}$, $b = 4r$. We expand $\left(\dfrac{1}{2r} - 4r\right)^2$:

$$= \left(\dfrac{1}{2r}\right)^2 - 2\left(\dfrac{1}{2r}\right)(4r) + (4r)^2$$

$$= \dfrac{1}{4r^2} - 4 + 16r^2$$

Using $(a + b + c)^2$ with $a = -3m$, $b = 4k$, $c = -l$. We expand $(-3m + 4k - l)^2$:

$$= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m)$$

$$= 9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm$$

Using $(a - b)^3$ with $a = x$, $b = \dfrac{1}{3}y$. We expand $\left(x - \dfrac{1}{3}y\right)^3$:

$$= x^3 - 3x^2\left(\dfrac{1}{3}y\right) + 3x\left(\dfrac{1}{3}y\right)^2 - \left(\dfrac{1}{3}y\right)^3$$

$$= x^3 - x^2y + \dfrac{1}{3}xy^2 - \dfrac{1}{27}y^3$$

Using $(a - b)^3$ with $a = \dfrac{7}{2}k$, $b = \dfrac{2}{3}m$. We expand $\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3$:

$$= \left(\dfrac{7}{2}k\right)^3 - 3\left(\dfrac{7}{2}k\right)^2\left(\dfrac{2}{3}m\right) + 3\left(\dfrac{7}{2}k\right)\left(\dfrac{2}{3}m\right)^2 - \left(\dfrac{2}{3}m\right)^3$$

$$= \dfrac{343}{8}k^3 - 3 \cdot \dfrac{49}{4} \cdot \dfrac{2}{3}k^2m + 3 \cdot \dfrac{7}{2} \cdot \dfrac{4}{9}km^2 - \dfrac{8}{27}m^3$$

$$= \dfrac{343}{8}k^3 - \dfrac{49}{2}k^2m + \dfrac{14}{3}km^2 - \dfrac{8}{27}m^3$$

17 and 21 are 2 less and 2 more than 19. Using $(a - b)(a + b) = a^2 - b^2$. We find $17 \times 21$:

$$= (19 - 2)(19 + 2)$$

$$= 19^2 - 2^2$$

$$= 361 - 4$$

$$= 357$$

We find $104 \times 96$:

$$= (100 + 4)(100 - 4)$$

$$= 100^2 - 4^2$$

$$= 10000 - 16$$

$$= 9984$$

We find $24 \times 16$:

$$= (20 + 4)(20 - 4)$$

$$= 20^2 - 4^2$$

$$= 400 - 16$$

$$= 384$$

Using $(a - b)^3$ with $147 = 150 - 3$. We find $147^3$:

$$= 150^3 - 3(150)^2(3) + 3(150)(3)^2 - 3^3$$

$$= 3375000 - 202500 + 4050 - 27$$

$$= 3176523$$

Using $(a - b)^3$ with $199 = 200 - 1$. We find $199^3$:

$$= 200^3 - 3(200)^2(1) + 3(200)(1)^2 - 1^3$$

$$= 8000000 - 120000 + 600 - 1$$

$$= 7880599$$

Using $(a - b)^3$ with $127 = 130 - 3$. We find $127^3$:

$$= 130^3 - 3(130)^2(3) + 3(130)(3)^2 - 3^3$$

$$= 2197000 - 152100 + 3510 - 27$$

$$= 2048383$$

The cube of a negative number is negative: $(-107)^3 = -(107)^3$. Using $(a + b)^3$ with $107 = 100 + 7$. We find $107^3$:

$$= 100^3 + 3(100)^2(7) + 3(100)(7)^2 + 7^3$$

$$= 1000000 + 210000 + 14700 + 343$$

$$= 1225043$$

∴ $(-107)^3 = $ $-1225043$.

$(-299)^3 = -(299)^3$. Using $(a - b)^3$ with $299 = 300 - 1$. We find $299^3$:

$$= 300^3 - 3(300)^2(1) + 3(300)(1)^2 - 1^3$$

$$= 27000000 - 270000 + 900 - 1$$

$$= 26730899$$

∴ $(-299)^3 = $ $-26730899$.

$4y^2 = (2y)^2$, $\dfrac{1}{16y^2} = \left(\dfrac{1}{4y}\right)^2$ and $2(2y)\left(\dfrac{1}{4y}\right) = 1$. We factor $4y^2 + 1 + \dfrac{1}{16y^2}$:

$$= (2y)^2 + 2(2y)\left(\dfrac{1}{4y}\right) + \left(\dfrac{1}{4y}\right)^2$$

$$= \left(2y + \dfrac{1}{4y}\right)^2$$

This is a difference of squares. We factor $9m^2 - \dfrac{1}{25n^2}$:

$$= (3m)^2 - \left(\dfrac{1}{5n}\right)^2$$

$$= \left(3m + \dfrac{1}{5n}\right)\left(3m - \dfrac{1}{5n}\right)$$

This is $a^3 - b^3$ with $a = 3b$ and the second term $\dfrac{1}{4b}$. We factor $27b^3 - \dfrac{1}{64b^3}$:

$$= (3b)^3 - \left(\dfrac{1}{4b}\right)^3$$

$$= \left(3b - \dfrac{1}{4b}\right)\left((3b)^2 + (3b)\left(\dfrac{1}{4b}\right) + \left(\dfrac{1}{4b}\right)^2\right)$$

$$= \left(3b - \dfrac{1}{4b}\right)\left(9b^2 + \dfrac{3}{4} + \dfrac{1}{16b^2}\right)$$

We need two numbers with sum $\dfrac{5}{6}$ and product $\dfrac{1}{6}$. They are $\dfrac{1}{2}$ and $\dfrac{1}{3}$ (since $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ and $\dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$). Using $(x + a)(x + b) = x^2 + (a + b)x + ab$:

$$x^2 + \dfrac{5x}{6} + \dfrac{1}{6} = \left(x + \dfrac{1}{2}\right)\left(x + \dfrac{1}{3}\right)$$

We rearrange the terms in falling powers of $u$: $27u^3 - \dfrac{27u^2}{5} + \dfrac{9u}{25} - \dfrac{1}{125}$. This matches $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a = 3u$, $b = \dfrac{1}{5}$:

$$3a^2b = 3(9u^2)\left(\dfrac{1}{5}\right) = \dfrac{27u^2}{5}$$

$$3ab^2 = 3(3u)\left(\dfrac{1}{25}\right) = \dfrac{9u}{25}$$

So:

$$27u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25} = \left(3u - \dfrac{1}{5}\right)^3$$

This is $a^3 + b^3$ with $a = 4y$, $b = \dfrac{z}{5}$. We factor $64y^3 + \dfrac{1}{125}z^3$:

$$= (4y)^3 + \left(\dfrac{z}{5}\right)^3$$

$$= \left(4y + \dfrac{z}{5}\right)\left((4y)^2 - (4y)\left(\dfrac{z}{5}\right) + \left(\dfrac{z}{5}\right)^2\right)$$

$$= \left(4y + \dfrac{z}{5}\right)\left(16y^2 - \dfrac{4yz}{5} + \dfrac{z^2}{25}\right)$$

This is $a^3 + b^3 + c^3 - 3abc$ with $a = p$, $b = 3q$, $c = r$, since $3abc = 3(p)(3q)(r) = 9pqr$. We factor $p^3 + 27q^3 + r^3 - 9pqr$:

$$= (p + 3q + r)\bigl(p^2 + (3q)^2 + r^2 - p(3q) - (3q)r - rp\bigr)$$

$$= (p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)$$

We factor $9m^2 - 12m + 4$:

$$= (3m)^2 - 2(3m)(2) + 2^2$$

$$= (3m - 2)^2$$

We take out $\dfrac{1}{3}$ so that each term is a perfect cube:

$$9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz = \dfrac{1}{3}(27x^3 - 8y^3 + z^3 + 18xyz)$$

Inside the bracket, $a = 3x$, $b = -2y$, $c = z$ gives $a^3 + b^3 + c^3 = 27x^3 - 8y^3 + z^3$ and $-3abc = -3(3x)(-2y)(z) = 18xyz$ ✓. Using the identity:

$$27x^3 - 8y^3 + z^3 + 18xyz = (3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)$$

∴

$$9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz = \dfrac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)$$

The squares are $(2x)^2$, $(3y)^2$ and $(6z)^2$, so we compare with $(2x + 3y + 6z)^2$:

$$(2x + 3y + 6z)^2 = 4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz$$

This has $12xy$ and $24xz$, but the printed question has $12xz$ and $24xy$ — the two numbers are swapped. This appears to be a misprint. The intended expression is $4x^2 + 9y^2 + 36z^2 + 24xz + 36yz + 12xy$, which factors as:

$$4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz = (2x + 3y + 6z)^2$$

(As printed, the expression cannot be a perfect square $(\pm 2x \pm 3y \pm 6z)^2$: such a square always has $xy$-term $\pm 12xy$ and $xz$-term $\pm 24xz$.)

In falling powers of $u$: $27u^3 - \dfrac{9u^2}{2} + \dfrac{u}{4} - \dfrac{1}{216}$. This matches $(a - b)^3$ with $a = 3u$, $b = \dfrac{1}{6}$:

$$3a^2b = 3(9u^2)\left(\dfrac{1}{6}\right) = \dfrac{9u^2}{2}$$

$$3ab^2 = 3(3u)\left(\dfrac{1}{36}\right) = \dfrac{u}{4}$$

So:

$$27u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4} = \left(3u - \dfrac{1}{6}\right)^3$$

$$4x^2 + 4x + 1 = (2x + 1)^2$$

$$4x^2 - 1 = (2x + 1)(2x - 1)$$

We simplify the expression:

$$= \dfrac{(2x + 1)^2}{(2x + 1)(2x - 1)}$$

$$= \dfrac{2x + 1}{2x - 1}$$

We factor the numerator:

$$9(3a^3 - 24b^3) = 27(a^3 - 8b^3)$$

$$9(3a^3 - 24b^3) = 27(a - 2b)(a^2 + 2ab + 4b^2)$$

We factor the denominator:

$$9a^2 - 36b^2 = 9(a^2 - 4b^2)$$

$$9a^2 - 36b^2 = 9(a - 2b)(a + 2b)$$

We simplify the expression:

$$= \dfrac{27(a - 2b)(a^2 + 2ab + 4b^2)}{9(a - 2b)(a + 2b)}$$

$$= \dfrac{3(a^2 + 2ab + 4b^2)}{a + 2b}$$

The numerator is $a^3 + b^3$ with $a = s$, $b = 5t$:

$$s^3 + 125t^3 = (s + 5t)(s^2 - 5st + 25t^2)$$

For the denominator, two numbers with sum $-2$ and product $-35$ are $-7$ and $5$:

$$s^2 - 2st - 35t^2 = (s - 7t)(s + 5t)$$

We simplify the expression:

$$= \dfrac{(s + 5t)(s^2 - 5st + 25t^2)}{(s - 7t)(s + 5t)}$$

$$= \dfrac{s^2 - 5st + 25t^2}{s - 7t}$$

The area of a rectangle is length × breadth, so we factor the area. We factor $25a^2 - 30ab + 9b^2$:

$$= (5a)^2 - 2(5a)(3b) + (3b)^2$$

$$= (5a - 3b)(5a - 3b)$$

∴ Length $= 5a - 3b$ units and breadth $= 5a - 3b$ units — the rectangle is a square.

We factor $36s^2 - 49t^2$:

$$= (6s)^2 - (7t)^2$$

$$= (6s + 7t)(6s - 7t)$$

∴ Length $= 6s + 7t$ units and breadth $= 6s - 7t$ units.

The volume of a cuboid is length × breadth × height, so we factor the volume into three factors. We factor $6a^2 - 24b^2$:

$$= 6(a^2 - 4b^2)$$

$$= 6(a + 2b)(a - 2b)$$

∴ One possible set: length $= a + 2b$, breadth $= a - 2b$, height $= 6$ units. (Other splits of the number 6, such as $2$, $3(a + 2b)$, $(a - 2b)$, are also correct.)

We take out the common factor $3p$, then find two numbers with sum $-5$ and product $4$: $-1$ and $-4$. We factor $3ps^2 - 15ps + 12p$:

$$= 3p(s^2 - 5s + 4)$$

$$= 3p(s - 1)(s - 4)$$

∴ One possible set: length $= 3p$, breadth $= s - 1$, height $= s - 4$ units.

The path goes all round the outside of the playground, so the outer square has side $40 + s + s = 40 + 2s$ metres.

The area of the path is the outer area minus the playground, so:

$$\text{Area of path} = (40 + 2s)^2 - 40^2$$

We simplify the area of the path:

$$= 40^2 + 2(40)(2s) + (2s)^2 - 40^2$$

$$= 1600 + 160s + 4s^2 - 1600$$

$$= 4s^2 + 160s$$

∴ The area of the path is $(4s^2 + 160s)$ square metres, that is, $4s(s + 40)$ m².

Let the number be $x$ ($x \neq 0$). The number plus its reciprocal is $\dfrac{10}{3}$, so:

$$x + \dfrac{1}{x} = \dfrac{10}{3}$$

Multiplying both sides by $3x$:

$$3x^2 + 3 = 10x$$

$$\Rightarrow 3x^2 - 10x + 3 = 0$$

We need two numbers with sum $-10$ and product $3 \times 3 = 9$: they are $-9$ and $-1$. We factor $3x^2 - 10x + 3$:

$$= 3x^2 - 9x - x + 3$$

$$= 3x(x - 3) - 1(x - 3)$$

$$= (x - 3)(3x - 1)$$

So $(x - 3)(3x - 1) = 0$:

$$\Rightarrow x - 3 = 0 \quad \text{or} \quad 3x - 1 = 0$$

$$\Rightarrow x = 3 \quad \text{or} \quad x = \dfrac{1}{3}$$

∴ The number is 3 or $\dfrac{1}{3}$. (Check: $3 + \dfrac{1}{3} = \dfrac{10}{3}$ ✓.)

Area = length × width, so we factor the area and look for the factor $2x + 1$. We need two numbers with sum $7$ and product $2 \times 3 = 6$: they are $6$ and $1$. We factor $2x^2 + 7x + 3$:

$$= 2x^2 + 6x + x + 3$$

$$= 2x(x + 3) + 1(x + 3)$$

$$= (2x + 1)(x + 3)$$$$\text{Length} = \dfrac{\text{Area}}{\text{Width}}$$

$$\text{Length} = \dfrac{(2x + 1)(x + 3)}{2x + 1}$$

$$\text{Length} = x + 3$$

∴ The length of the pool is $(x + 3)$ hastas.

$px^2 + 5x + r$ has degree 2, and $x - 2$ and $x - \dfrac{1}{2}$ are two linear factors, so the remaining factor is just the number $p$ (the coefficient of $x^2$):

$$px^2 + 5x + r = p(x - 2)\left(x - \dfrac{1}{2}\right)$$

We expand the right-hand side:

$$= p\left(x^2 - \dfrac{1}{2}x - 2x + 1\right)$$

$$= p\left(x^2 - \dfrac{5}{2}x + 1\right)$$

$$= px^2 - \dfrac{5p}{2}x + p$$

So:

$$px^2 + 5x + r = px^2 - \dfrac{5p}{2}x + p$$

This is an identity, so the constant terms on both sides are equal:

$$\Rightarrow r = p$$

∴ $p = r$. (Comparing the $x$ terms also gives $5 = -\dfrac{5p}{2}$, so $p = r = -2$.)

We are given:

$$a + b + c = 5 \quad \cdots (1)$$

$$ab + bc + ca = 10 \quad \cdots (2)$$

First we find $a^2 + b^2 + c^2$ using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$.

Substituting equation (1) and equation (2):

$$5^2 = a^2 + b^2 + c^2 + 2(10)$$

$$\Rightarrow a^2 + b^2 + c^2 = 25 - 20$$

$$\Rightarrow a^2 + b^2 + c^2 = 5 \quad \cdots (3)$$

Now we use $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Substituting equation (1), equation (2) and equation (3):

$$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10)$$

$$a^3 + b^3 + c^3 - 3abc = 5 \times (-5)$$

$$a^3 + b^3 + c^3 - 3abc = -25$$

Hence proved.

We factor $n^3 - n$:

$$= n(n^2 - 1)$$

$$= n(n - 1)(n + 1)$$

$$= (n - 1)\,n\,(n + 1)$$

So $n^3 - n$ is the product of three consecutive integers.

• Among any two consecutive integers, one is even, so the product is divisible by 2.
• Among any three consecutive integers, one is a multiple of 3, so the product is divisible by 3.

A number divisible by both 2 and 3 (which have no common factor) is divisible by $2 \times 3 = 6$.

∴ $n^3 - n$ is always divisible by 6. For example, $n = 5$: $5^3 - 5 = 120 = 6 \times 20$.

We write the expression as $a^3 + b^3 + c^3 - 3abc$ with $a = x$, $b = y$, $c = 4$:

$$x^3 + y^3 + 4^3 - 3(x)(y)(4) = x^3 + y^3 + 64 - 12xy$$

which is exactly our expression. By the identity:

$$x^3 + y^3 - 12xy + 64 = (x + y + 4)(x^2 + y^2 + 16 - xy - 4y - 4x)$$

We are given $x + y = -4$:

$$\Rightarrow x + y + 4 = 0$$

So the first factor is 0, and:

$$x^3 + y^3 - 12xy + 64 = 0 \times (x^2 + y^2 + 16 - xy - 4y - 4x)$$

$$x^3 + y^3 - 12xy + 64 = 0$$

∴ The value is 0.

We write the expression as $a^3 + b^3 + c^3 - 3abc$ with $a = x$, $b = -2y$, $c = -6$:

$$a^3 + b^3 + c^3 = x^3 - 8y^3 - 216$$

$$-3abc = -3(x)(-2y)(-6) = -36xy$$

which matches our expression. By the identity:

$$x^3 - 8y^3 - 36xy - 216 = (x - 2y - 6)(x^2 + 4y^2 + 36 + 2xy - 12y + 6x)$$

We are given $x = 2y + 6$:

$$\Rightarrow x - 2y - 6 = 0$$

So the first factor is 0, and:

$$x^3 - 8y^3 - 36xy - 216 = 0$$

∴ The value is 0.