Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Fill in the blanks to complete the following identities:
(i) $s^2-11s+24=(\_\_\_\_)(\_\_\_\_)$
Solution:
We have to factorise the quadratic expression
$$ s^2-11s+24. $$We need two numbers whose:
- Product is $24$, and
- Sum is $-11$.
The required numbers are
$$ -3 \quad \text{and} \quad -8, $$because
$$ (-3)\times(-8)=24 $$and
$$ (-3)+(-8)=-11. $$Hence,
$$ \boxed{s^2-11s+24=(s-3)(s-8)} $$(ii) $(\_\_\_\_)(x+1)=3x^2-4x-7$
Solution:
We have to find the missing factor.
Factorise the given quadratic expression.
$$ 3x^2-4x-7. $$Find two numbers whose product is
$$ 3\times(-7)=-21 $$and whose sum is
$$ -4. $$The required numbers are
$$ -7 \quad \text{and} \quad 3. $$Split the middle term.
$$ \begin{aligned} 3x^2-4x-7 &=3x^2-7x+3x-7\\ &=x(3x-7)+1(3x-7)\\ &=(x+1)(3x-7). \end{aligned} $$Therefore, the missing factor is
$$ \boxed{3x-7.} $$(iii) $10x^2-11x-6=(2x-\_\_)(\_\_+2)$
Solution:
Factorise the given quadratic expression.
$$ 10x^2-11x-6. $$Find two numbers whose product is
$$ 10\times(-6)=-60 $$and whose sum is
$$ -11. $$The required numbers are
$$ -15 \quad \text{and} \quad 4. $$Split the middle term.
$$ \begin{aligned} 10x^2-11x-6 &=10x^2-15x+4x-6\\ &=5x(2x-3)+2(2x-3)\\ &=(2x-3)(5x+2). \end{aligned} $$Hence, the blanks are
$$ \boxed{3 \quad \text{and} \quad 5x.} $$Therefore,
$$ \boxed{10x^2-11x-6=(2x-3)(5x+2)} $$(iv) $6x^2+7x+2=(\_\_\_\_)(\_\_\_\_)$
Solution:
Factorise the given quadratic expression.
$$ 6x^2+7x+2. $$Find two numbers whose product is
$$ 6\times2=12 $$and whose sum is
$$ 7. $$The required numbers are
$$ 3 \quad \text{and} \quad 4. $$Split the middle term.
$$ \begin{aligned} 6x^2+7x+2 &=6x^2+3x+4x+2\\ &=3x(2x+1)+2(2x+1)\\ &=(3x+2)(2x+1). \end{aligned} $$Answer:
$$ \boxed{6x^2+7x+2=(3x+2)(2x+1)} $$Question 2. Select and use the identity that will help you to find the following products without multiplying directly.
(i) $(41)^2$
Solution:
Since $41$ is close to $40$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $41$ as
$$ 41=40+1. $$Here,
$$ a=40 \qquad\text{and}\qquad b=1. $$Using the identity,
$$ \begin{aligned} 41^2 &=(40+1)^2\\ &=40^2+2(40)(1)+1^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 40^2&=1600,\\[4pt] 2(40)(1)&=80,\\[4pt] 1^2&=1. \end{aligned} $$Add these values.
$$ \begin{aligned} 41^2 &=1600+80+1\\ &=1681. \end{aligned} $$Answer:
$$ \boxed{41^2=1681} $$(ii) $(27)^2$
Solution:
Since $27$ is close to $30$, we use the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Express $27$ as
$$ 27=30-3. $$Here,
$$ a=30 \qquad\text{and}\qquad b=3. $$Using the identity,
$$ \begin{aligned} 27^2 &=(30-3)^2\\ &=30^2-2(30)(3)+3^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 30^2&=900,\\[4pt] 2(30)(3)&=180,\\[4pt] 3^2&=9. \end{aligned} $$Add these values.
$$ \begin{aligned} 27^2 &=900-180+9\\ &=729. \end{aligned} $$Answer:
$$ \boxed{27^2=729} $$(iii) $(23\times17)$
Solution:
Observe that
$$ 23=20+3 \qquad\text{and}\qquad 17=20-3. $$Since the numbers are of the form $(a+b)$ and $(a-b)$, we use the identity
$$ (a+b)(a-b)=a^2-b^2. $$Here,
$$ a=20 \qquad\text{and}\qquad b=3. $$Using the identity,
$$ \begin{aligned} 23\times17 &=(20+3)(20-3)\\ &=20^2-3^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 20^2&=400,\\[4pt] 3^2&=9. \end{aligned} $$Subtract the values.
$$ \begin{aligned} 23\times17 &=400-9\\ &=391. \end{aligned} $$Answer:
$$ \boxed{23\times17=391} $$(iv) $(135)^2$
Solution:
Since $135$ is close to $100$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $135$ as
$$ 135=100+35. $$Here,
$$ a=100 \qquad\text{and}\qquad b=35. $$Using the identity,
$$ \begin{aligned} 135^2 &=(100+35)^2\\ &=100^2+2(100)(35)+35^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^2&=10000,\\[4pt] 2(100)(35)&=7000,\\[4pt] 35^2&=1225. \end{aligned} $$Add these values.
$$ \begin{aligned} 135^2 &=10000+7000+1225\\ &=18225. \end{aligned} $$Answer:
$$ \boxed{135^2=18225} $$(v) $(97)^2$
Solution:
Since $97$ is close to $100$, we use the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Express $97$ as
$$ 97=100-3. $$Here,
$$ a=100 \qquad\text{and}\qquad b=3. $$Using the identity,
$$ \begin{aligned} 97^2 &=(100-3)^2\\ &=100^2-2(100)(3)+3^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^2&=10000,\\[4pt] 2(100)(3)&=600,\\[4pt] 3^2&=9. \end{aligned} $$Add these values.
$$ \begin{aligned} 97^2 &=10000-600+9\\ &=9409. \end{aligned} $$Answer:
$$ \boxed{97^2=9409} $$(vi) $(18\times29)$
Solution:
Observe that
$$ 18=23-5 \qquad\text{and}\qquad 29=23+6. $$A simpler approach is to write
$$ 18=20-2 \qquad\text{and}\qquad 29=20+9. $$Now use the identity
$$ (x+a)(x+b)=x^2+x(a+b)+ab. $$Here,
$$ x=20,\qquad a=-2,\qquad b=9. $$Using the identity,
$$ \begin{aligned} 18\times29 &=(20-2)(20+9)\\ &=20^2+20(-2+9)+(-2)(9). \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 20^2&=400,\\[4pt] 20(-2+9)&=20\times7=140,\\[4pt] (-2)(9)&=-18. \end{aligned} $$Therefore,
$$ \begin{aligned} 18\times29 &=400+140-18\\ &=522. \end{aligned} $$Answer:
$$ \boxed{18\times29=522} $$(vii) $(34\times43)$
Solution:
Observe that
$$ 34=40-6 \qquad\text{and}\qquad 43=40+3. $$Use the identity
$$ (x+a)(x+b)=x^2+x(a+b)+ab. $$Here,
$$ x=40,\qquad a=-6,\qquad b=3. $$Using the identity,
$$ \begin{aligned} 34\times43 &=(40-6)(40+3)\\ &=40^2+40(-6+3)+(-6)(3). \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 40^2&=1600,\\[4pt] 40(-6+3)&=40(-3)=-120,\\[4pt] (-6)(3)&=-18. \end{aligned} $$Therefore,
$$ \begin{aligned} 34\times43 &=1600-120-18\\ &=1462. \end{aligned} $$Answer:
$$ \boxed{34\times43=1462} $$(viii) $(205)^2$
Solution:
Since $205$ is close to $200$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $205$ as
$$ 205=200+5. $$Here,
$$ a=200 \qquad\text{and}\qquad b=5. $$Using the identity,
$$ \begin{aligned} 205^2 &=(200+5)^2\\ &=200^2+2(200)(5)+5^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^2&=40000,\\[4pt] 2(200)(5)&=2000,\\[4pt] 5^2&=25. \end{aligned} $$Therefore,
$$ \begin{aligned} 205^2 &=40000+2000+25\\ &=42025. \end{aligned} $$Answer:
$$ \boxed{205^2=42025} $$Question 3. Factor the following:
(i) $9a^2+b^2+4c^2-6ab+12ac-4bc$
Solution:
We know the identity
$$ (x-y+z)^2=x^2+y^2+z^2-2xy+2xz-2yz. $$Compare the given expression with the identity.
The square terms are
$$ 9a^2=(3a)^2, $$ $$ b^2=b^2, $$ $$ 4c^2=(2c)^2. $$Now check the remaining terms.
$$ \begin{aligned} -2(3a)(b) &=-6ab, \\[6pt] 2(3a)(2c) &=12ac, \\[6pt] -2(b)(2c) &=-4bc. \end{aligned} $$These are exactly the middle terms of the given expression.
Hence,
$$ \begin{aligned} 9a^2+b^2+4c^2-6ab+12ac-4bc &=(3a-b+2c)^2. \end{aligned} $$Answer:
$$ \boxed{(3a-b+2c)^2} $$(ii) $16s^2+25t^2-40st$
Solution:
We know the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Compare the given expression with the identity.
$$ 16s^2=(4s)^2, $$ $$ 25t^2=(5t)^2, $$and
$$ 2(4s)(5t)=40st. $$Therefore, the given expression is of the form
$$ a^2-2ab+b^2, $$where
$$ a=4s, \qquad b=5t. $$Hence,
$$ \begin{aligned} 16s^2+25t^2-40st &=(4s-5t)^2. \end{aligned} $$Answer:
$$ \boxed{(4s-5t)^2} $$(iii) $r^2-r-42$
Solution:
We have to factorise
$$ r^2-r-42. $$Find two numbers whose
- Product is $-42$, and
- Sum is $-1$.
The required numbers are
$$ -7 \qquad\text{and}\qquad 6, $$because
$$ (-7)\times6=-42 $$and
$$ (-7)+6=-1. $$Split the middle term.
$$ \begin{aligned} r^2-r-42 &=r^2-7r+6r-42. \end{aligned} $$Group the terms.
$$ \begin{aligned} &=r(r-7)+6(r-7). \end{aligned} $$Take the common factor $(r-7)$.
$$ \begin{aligned} &=(r+6)(r-7). \end{aligned} $$Answer:
$$ \boxed{(r+6)(r-7)} $$(iv) $49g^2+14gh+h^2$
Solution:
We know the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
The square terms are
$$ 49g^2=(7g)^2, $$ $$ h^2=h^2. $$Now check the middle term.
$$ \begin{aligned} 2(7g)(h) &=14gh. \end{aligned} $$This is exactly the middle term of the given expression.
Therefore, the given expression is of the form
$$ a^2+2ab+b^2, $$where
$$ a=7g, \qquad b=h. $$Using the identity, we obtain
$$ \begin{aligned} 49g^2+14gh+h^2 &=(7g+h)^2. \end{aligned} $$Answer:
$$ \boxed{(7g+h)^2} $$(v) $64u^2+121v^2+4w^2-176uv-32uw+44vw$
Solution:
We know the identity
$$ (a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc. $$Compare the given expression with the identity.
The square terms are
$$ 64u^2=(8u)^2, $$ $$ 121v^2=(11v)^2, $$ $$ 4w^2=(2w)^2. $$Now check the remaining terms.
$$ \begin{aligned} -2(8u)(11v) &=-176uv, \\[6pt] -2(8u)(2w) &=-32uw, \\[6pt] 2(11v)(2w) &=44vw. \end{aligned} $$These are exactly the remaining terms of the given expression.
Therefore, the given expression is of the form
$$ (a-b-c)^2, $$where
$$ a=8u, \qquad b=11v, \qquad c=2w. $$Using the identity, we get
$$ \begin{aligned} 64u^2+121v^2+4w^2-176uv-32uw+44vw &=(8u-11v-2w)^2. \end{aligned} $$Verification:
Expanding $(8u-11v-2w)^2$ gives
$$ \begin{aligned} &(8u)^2+(11v)^2+(2w)^2\\ &\quad-2(8u)(11v)\\ &\quad-2(8u)(2w)\\ &\quad+2(11v)(2w)\\[4pt] &=64u^2+121v^2+4w^2\\ &\quad-176uv-32uw+44vw. \end{aligned} $$This is exactly the given expression.
Answer:
$$ \boxed{(8u-11v-2w)^2} $$