Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) $117^2$
Solution:
Since $117$ is close to $100$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $117$ as
$$ 117=100+17. $$Here,
$$ a=100 \qquad\text{and}\qquad b=17. $$Using the identity,
$$ \begin{aligned} 117^2 &=(100+17)^2\\ &=100^2+2(100)(17)+17^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 100^2&=10000,\\[4pt] 2(100)(17)&=3400,\\[4pt] 17^2&=289. \end{aligned} $$Adding these values,
$$ \begin{aligned} 117^2 &=10000+3400+289\\ &=13689. \end{aligned} $$Answer:
$$ \boxed{117^2=13689} $$(ii) $78^2$
Solution:
Since $78$ is close to $80$, we use the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Express $78$ as
$$ 78=80-2. $$Here,
$$ a=80 \qquad\text{and}\qquad b=2. $$Using the identity,
$$ \begin{aligned} 78^2 &=(80-2)^2\\ &=80^2-2(80)(2)+2^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 80^2&=6400,\\[4pt] 2(80)(2)&=320,\\[4pt] 2^2&=4. \end{aligned} $$Adding these values,
$$ \begin{aligned} 78^2 &=6400-320+4\\ &=6084. \end{aligned} $$Answer:
$$ \boxed{78^2=6084} $$(iii) $198^2$
Solution:
Since $198$ is close to $200$, we use the identity
$$ (a-b)^2=a^2-2ab+b^2. $$Express $198$ as
$$ 198=200-2. $$Here,
$$ a=200 \qquad\text{and}\qquad b=2. $$Using the identity,
$$ \begin{aligned} 198^2 &=(200-2)^2\\ &=200^2-2(200)(2)+2^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^2&=40000,\\[4pt] 2(200)(2)&=800,\\[4pt] 2^2&=4. \end{aligned} $$Adding these values,
$$ \begin{aligned} 198^2 &=40000-800+4\\ &=39204. \end{aligned} $$Answer:
$$ \boxed{198^2=39204} $$(iv) $214^2$
Solution:
Since $214$ is close to $200$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $214$ as
$$ 214=200+14. $$Here,
$$ a=200 \qquad\text{and}\qquad b=14. $$Using the identity,
$$ \begin{aligned} 214^2 &=(200+14)^2\\ &=200^2+2(200)(14)+14^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^2&=40000,\\[4pt] 2(200)(14)&=5600,\\[4pt] 14^2&=196. \end{aligned} $$Adding these values,
$$ \begin{aligned} 214^2 &=40000+5600+196\\ &=45796. \end{aligned} $$Answer:
$$ \boxed{214^2=45796} $$(v) $1104^2$
Solution:
Since $1104$ is close to $1100$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $1104$ as
$$ 1104=1100+4. $$Here,
$$ a=1100 \qquad\text{and}\qquad b=4. $$Using the identity,
$$ \begin{aligned} 1104^2 &=(1100+4)^2\\ &=1100^2+2(1100)(4)+4^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 1100^2&=1210000,\\[4pt] 2(1100)(4)&=8800,\\[4pt] 4^2&=16. \end{aligned} $$Adding these values,
$$ \begin{aligned} 1104^2 &=1210000+8800+16\\ &=1218816. \end{aligned} $$Answer:
$$ \boxed{1104^2=1218816} $$(vi) $1120^2$
Solution:
Since $1120$ is close to $1100$, we use the identity
$$ (a+b)^2=a^2+2ab+b^2. $$Express $1120$ as
$$ 1120=1100+20. $$Here,
$$ a=1100 \qquad\text{and}\qquad b=20. $$Using the identity,
$$ \begin{aligned} 1120^2 &=(1100+20)^2\\ &=1100^2+2(1100)(20)+20^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 1100^2&=1210000,\\[4pt] 2(1100)(20)&=44000,\\[4pt] 20^2&=400. \end{aligned} $$Adding these values,
$$ \begin{aligned} 1120^2 &=1210000+44000+400\\ &=1254400. \end{aligned} $$Answer:
$$ \boxed{1120^2=1254400} $$Question 2. Factor using suitable identities:
(i) $16y^2-24y+9$
Solution:
We know that
$$ (a-b)^2=a^2-2ab+b^2. $$Compare the given expression with the identity.
$$ 16y^2-24y+9 $$Here,
$$ 16y^2=(4y)^2, $$ $$ 9=3^2, $$and
$$ 2(4y)(3)=24y. $$Thus, the given expression is of the form $a^2-2ab+b^2$, where
$$ a=4y \qquad\text{and}\qquad b=3. $$Using the identity, we get
$$ \begin{aligned} 16y^2-24y+9 &=(4y-3)^2. \end{aligned} $$Answer:
$$ \boxed{(4y-3)^2} $$(ii) $\dfrac94s^2+6st+4t^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
$$ \frac94s^2+6st+4t^2 $$Here,
$$ \frac94s^2=\left(\frac32s\right)^2, $$ $$ 4t^2=(2t)^2, $$and
$$ 2\left(\frac32s\right)(2t)=6st. $$Thus, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=\frac32s \qquad\text{and}\qquad b=2t. $$Using the identity, we get
$$ \begin{aligned} \frac94s^2+6st+4t^2 &=\left(\frac32s+2t\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(\frac32s+2t\right)^2} $$(iii) $\dfrac{m^2}{9}+\dfrac{mk}{3}+\dfrac{k^2}{4}+3nk+2mn+9n^2$
Solution:
We know that
$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. $$Compare the given expression with the identity.
The square terms are
$$ \frac{m^2}{9}=\left(\frac{m}{3}\right)^2, $$ $$ \frac{k^2}{4}=\left(\frac{k}{2}\right)^2, $$ $$ 9n^2=(3n)^2. $$Now check the remaining terms.
$$ \begin{aligned} 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) &=\frac{mk}{3}, \\[6pt] 2\left(\frac{m}{3}\right)(3n) &=2mn, \\[6pt] 2\left(\frac{k}{2}\right)(3n) &=3nk. \end{aligned} $$These are exactly the remaining terms of the given expression.
Therefore, the given expression is of the form
$$ (a+b+c)^2, $$where
$$ a=\frac{m}{3}, \qquad b=\frac{k}{2}, \qquad c=3n. $$Using the identity, we obtain
$$ \begin{aligned} \frac{m^2}{9} +\frac{mk}{3} +\frac{k^2}{4} +3nk +2mn +9n^2 &= \left(\frac{m}{3}+\frac{k}{2}+3n\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(\frac{m}{3}+\frac{k}{2}+3n\right)^2} $$(iv) $\dfrac{p^2}{16}-2+\dfrac{16}{p^2}$
Solution:
We know that
$$ (a-b)^2=a^2-2ab+b^2. $$Compare the given expression with the identity.
$$ \frac{p^2}{16}-2+\frac{16}{p^2} $$Here,
$$ \frac{p^2}{16}=\left(\frac{p}{4}\right)^2, $$ $$ \frac{16}{p^2}=\left(\frac{4}{p}\right)^2. $$Now check the middle term.
$$ \begin{aligned} 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) &=2\times1\\ &=2. \end{aligned} $$Hence, the given expression is of the form $a^2-2ab+b^2$, where
$$ a=\frac{p}{4} \qquad\text{and}\qquad b=\frac{4}{p}. $$Using the identity, we obtain
$$ \begin{aligned} \frac{p^2}{16}-2+\frac{16}{p^2} &=\left(\frac{p}{4}-\frac{4}{p}\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(\frac{p}{4}-\frac{4}{p}\right)^2} $$(v) $9a^2+4b^2+c^2-12ab+6ac-4bc$
Solution:
We know that
$$ (a-b+c)^2=a^2+b^2+c^2-2ab+2ac-2bc. $$Compare the given expression with the identity.
The square terms are
$$ 9a^2=(3a)^2, $$ $$ 4b^2=(2b)^2, $$ $$ c^2=c^2. $$Now check the remaining terms.
$$ \begin{aligned} -2(3a)(2b) &=-12ab, \\[6pt] 2(3a)(c) &=6ac, \\[6pt] -2(2b)(c) &=-4bc. \end{aligned} $$These are exactly the remaining terms of the given expression.
Therefore, the given expression is of the form
$$ (a-b+c)^2, $$where
$$ a=3a, \qquad b=2b, \qquad c=c. $$Using the identity, we get
$$ \begin{aligned} 9a^2+4b^2+c^2-12ab+6ac-4bc &=(3a-2b+c)^2. \end{aligned} $$Answer:
$$ \boxed{(3a-2b+c)^2} $$Question 3. Expand the following using the identity $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$.
(i) $(p+3q+7r)^2$
Solution:
We know that
$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. $$Here,
$$ a=p, \qquad b=3q, \qquad c=7r. $$Substitute these values into the identity.
$$ \begin{aligned} (p+3q+7r)^2 &=p^2+(3q)^2+(7r)^2\\ &\quad+2(p)(3q)+2(3q)(7r)+2(p)(7r). \end{aligned} $$Now simplify each term.
$$ \begin{aligned} p^2&=p^2,\\[4pt] (3q)^2&=9q^2,\\[4pt] (7r)^2&=49r^2,\\[4pt] 2(p)(3q)&=6pq,\\[4pt] 2(3q)(7r)&=42qr,\\[4pt] 2(p)(7r)&=14pr. \end{aligned} $$Substituting these values, we get
$$ \begin{aligned} (p+3q+7r)^2 &=p^2+9q^2+49r^2\\ &\quad+6pq+14pr+42qr. \end{aligned} $$Answer:
$$ \boxed{ (p+3q+7r)^2 = p^2+9q^2+49r^2+6pq+14pr+42qr } $$(ii) $(3x-2y+4z)^2$
Solution:
We know that
$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. $$Here,
$$ a=3x, \qquad b=-2y, \qquad c=4z. $$Substitute these values into the identity.
$$ \begin{aligned} (3x-2y+4z)^2 &=(3x)^2+(-2y)^2+(4z)^2\\ &\quad+2(3x)(-2y)\\ &\quad+2(-2y)(4z)\\ &\quad+2(3x)(4z). \end{aligned} $$Now simplify each term.
$$ \begin{aligned} (3x)^2&=9x^2,\\[4pt] (-2y)^2&=4y^2,\\[4pt] (4z)^2&=16z^2,\\[4pt] 2(3x)(-2y)&=-12xy,\\[4pt] 2(-2y)(4z)&=-16yz,\\[4pt] 2(3x)(4z)&=24xz. \end{aligned} $$Substituting these values, we get
$$ \begin{aligned} (3x-2y+4z)^2 &=9x^2+4y^2+16z^2\\ &\quad-12xy+24xz-16yz. \end{aligned} $$Answer:
$$ \boxed{ (3x-2y+4z)^2 = 9x^2+4y^2+16z^2-12xy+24xz-16yz } $$Question 4. Is this an identity?
$$ (a+b-c)^2+(a-b+c)^2+(a-b-c)^2=2a^2+2b^2+2c^2 $$Solution:
To check whether the given equation is an identity, we simplify the left-hand side and compare it with the right-hand side.
Using the identity
$$ (x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx, $$we expand each expression separately.
First expression:
$$ \begin{aligned} (a+b-c)^2 &=a^2+b^2+c^2+2ab-2ac-2bc. \end{aligned} $$Second expression:
$$ \begin{aligned} (a-b+c)^2 &=a^2+b^2+c^2-2ab+2ac-2bc. \end{aligned} $$Third expression:
$$ \begin{aligned} (a-b-c)^2 &=a^2+b^2+c^2-2ab-2ac+2bc. \end{aligned} $$Now add the three expressions.
$$ \begin{aligned} &(a+b-c)^2+(a-b+c)^2+(a-b-c)^2\\[4pt] &=(a^2+b^2+c^2+2ab-2ac-2bc)\\ &\quad+(a^2+b^2+c^2-2ab+2ac-2bc)\\ &\quad+(a^2+b^2+c^2-2ab-2ac+2bc). \end{aligned} $$Combine the like terms.
$$ \begin{aligned} &=3a^2+3b^2+3c^2\\ &\quad+(2ab-2ab-2ab)\\ &\quad+(-2ac+2ac-2ac)\\ &\quad+(-2bc-2bc+2bc). \end{aligned} $$Simplifying further,
$$ \begin{aligned} &=3a^2+3b^2+3c^2-2ab-2ac-2bc. \end{aligned} $$The right-hand side is
$$ 2a^2+2b^2+2c^2. $$Since
$$ 3a^2+3b^2+3c^2-2ab-2ac-2bc \neq 2a^2+2b^2+2c^2, $$the left-hand side is not equal to the right-hand side for all values of $a$, $b$, and $c$.
Therefore, the given equation is not an identity.
Answer:
$$ \boxed{\text{No, the given equation is not an identity.}} $$