Exercise 2.5

Q1. Use suitable identities to find the following products :

i) $(x+4)(x+10)$ ii) $(x+8)(x-10)$

iii) $(3x+4)(3x-5)$ iv) $(y^2+frac(3)(2))(y^2-frac(3)(2))$

v) $(3-2x)(3+2x)$

Sol. : i)

$(x+a)(x+b)=x^2+(a+b)x+ab$ [We use an identity]

$(x+4)(x+10)=x^2+(4+10)x+4×10$

$(x+4)(x+10)=x^2+14x+40$

Answer :

$=x^2+14x+40$

ii)

$(x+a)(x-b)=x^2+(a-b)x-ab$ [We use an identity]

$(x+8)(x-10)=x^2+(8-10)x-8×10$

$(x+8)(x-10)=x^2-2x-80$

Answer :

$=x^2-2x-80$

iii)

$(x+a)(x-b)=x^2+(a-b)x-ab$ [We use an identity]

$(3x+4)(3x-5)=(3x)^2+(4-5)3x-4×5$

$(3x+4)(3x-5)=9x^2-3x-20$

Answer :

$=9x^2-3x-20$

iv)

$(a+b)(a-b)=a^2-b^2$ [We use an identity]

$(y^2 + frac(3)(2))(y^2 - frac(3)(2))=(y^2)^2-(frac(3)(2))^2$

$(y^2 + frac(3)(2))(y^2 - frac(3)(2))=y^4-frac(9)(4)$

Answer :

$=y^4-frac(9)(4)$

$(a-b)(a+b)=a^2-b^2$ [We use an identity]

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

$(3-2x)(3+2x)=9-4x^2$

Answer :

$=9-4x^2$

Q2.Evaluate the following products without multiplying directly :

i) $103×107$ ii) $95×96$ iii) $104×96$

Sol. :

$(100+3)(100+7)$

$(x+a)(x+b)=x^2+(a+b)x+ab$ [We use an identity]

Where

$x=100, a=3, and b=7$

$(100+3)(100+7)=100^2+(3+7)100+3×7$

$(100+3)(100+7)=10000+1000+21$

$(100+3)(100+7)=11021$

Answer :

$=11021$

ii)

$(100-5)(100-4)$

$(x-a)(x-b)=x^2-(a+b)x+ab$ [We use an identity]

Where

$x=100, a=3$ , and

$b=7$

$(100-5)(100-4)=100^2-(5+4)100+5×4$

$(100-5)(100-4)=10000-900+20$

$(100-5)(100-4)=9120$

Answer :

$=9120$

iii)

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

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

Where

$a=100, b=4$

$(100+4)(100-4)=(100)^2-(4)^2$

$(100+4)(100-4)=10000-16$

$(100+4)(100-4)=9984$

Answer :

$=9984$

Q3. Factorise the following using appropriate identities :

i) $9x^2+6xy+y^2$ ii) $4y^2-4y+1$ iii) $x^2 - frac(y^2)(100)$

Sol. :

$9x^2+6xy+y^2$

$=(3x)^2+2×3xy+(y)^2$

$a^2+2ab+b^2=(a+b)^2$ [We use an identity]

Where

$a=3$ and

$b=y$

$(3x)^2+2×3xy+(y)^2=(3x+y)^2$

Answer :

$=(3x+y)^2$

ii)

$4y^2-4y+1$

$=(2y)^2-2×2y×1+(1)^2$

$a^2-2ab+b^2=(a-b)^2$ [We use an identity]

Where

$a=2y$ and

$b=1$

$(2y)^2-2×2y×1+(1)^2=(2y-1)^2$

Answer :

$=(2y-1)^2$

iii)

$(x)^2-frac(y^2)(100)$

$=(x)^2-(frac(y)(10))^2$

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

Where

$a=x$ and

$b=frac(y)(10)$

$(x)^2-(frac(y)(10))^2=(x-frac(y)(10))(x+frac(y)(10))$

Answer :

$=(x-frac(y)(10))(x+frac(y)(10))$

Q4.Expand each of the following using suitable identity :

i) $(x+2y+4z)^2$ ii) $(2x-y+z)^2$

iii) $(-2x+3y+2z)^2$ iv) $( 3a-7b-c)^2$

v) $(-2x+5y-3z)^2$ vi) $[frac(1)(4)a-frac(1)(2)b+1]$

Sol. :

$(x+2x+4z)^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where

$p=x, q=2x$ and

$r=4z$

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

$+2(x×2x+2x×4z+4z×x)$

$(x+2x+4z)^2=x^2+4x^2+16z^2$

$+2(2x^2+8xz+4zx)$

$(x+2x+4z)^2=x^2+4x^2+16z^2$

$+4x^2+16xz+8zx$

Answer :

$=x^2+4x^2+16z^2+2x^2+16xz+8zx$

ii)

$(2x-y+z)^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where

$p=2x, q=-y$ and

$r=z$

$(2x-y+z)^2=(2x)^2+(-y)^2+(z)^2$

$+2[2x(-y)+(-y)z+z2x]$

$(2x-y+z)^2=4x^2+y^2+z^2$

$+2[-2xy-yz+2xz]$

$(2x-y+z)^2=4x^2+y^2+z^2$

$-4xy-2yz+4xz$

Answer :

$=4x^2+y^2+z^2-4xy-2yz+4xz$

iii)

$(-2x+3y+2z)^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where $p=(-2x), q=3y,$ and $r=2z$

$(-2x+3y+2z)^2=(-2x)^2+(3y)^2+(2z)^2$

$+2[(-2x)(3y)+(3y)(2z)+(2z)(-2x)]$

$(-2x+3y+2z)^z=4x^2+9y^2+4z^2$

$+2[-6xy+6yz-4xz]$

$(-2x+3y+2z)^2=4x^2+9y^2+4z^2$

$-12xy+12yz-8xz$

Answer :

$=4x^2+9y^2+4z^2-12xy+12yz-8xz$

iv)

$( 3a-7b-c)^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where $p=3a, q=-7b$ and $r=-c$

$(3a-7b-c)^2=(3a)^2+(-7b)^2+(-c)^2$

$+2[(3a)(-7b)+(-7b)(-c)+(-c)(3a)]$

$(3a-7b-c)^2=9a^2+49b^2-c^2$

$+2[-21ab+7bc-3ac]$

$(3a-7b-c)^2=9a^2+49b^2-c^2-42ab+14bc-6ac$

Answer :

$=9a^2+49b^2-c^2-42ab+14bc-6ac$

$(-2x+5y-3z)^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where $p=-2x,q=5y,$ and $r=-3z$

$(-2x+5y-3z)^2=(-2x)^2+(5y)^2+(-3z)^2$

$+2[(-2x)(5y)+(5y)(-3z)+(-3z)(-2x)]$

$(-2x+5y-3z)^2=4x^2+25y^2+9z^2$

$+2[-10xy-15yz+6xz]$

$(-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy$

$-30yz+12xz$

Answer :

$=4x^2+25y^2+9z^2-20xy-30yz+12xz$

vi)

$[frac(1)(4)a-frac(1)(2)b+1]^2$

$(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$ [We use an identity]

Where $p=frac(1)(4), q=(-frac(1)(2))$ and $r=1$

$[frac(1)(4)a-frac(1)(2)b+1]^2=(afrac(1)(4))^2+(-bfrac(1)(2)^2)+(1)^2$

$+2[(afrac(1)(4))(-bfrac(1)(2))+(-bfrac(1)(2))(1)+(1)(afrac(1)(4))]$

$[frac(1)(4)a-frac(1)(2)b+1]^2=a^2frac(1)(16)+b^2frac(1)(4)+1$

$+2[-abfrac(1)(8)-bfrac(1)(2)+afrac(1)(4)]$

$[frac(1)(4)a-frac(1)(2)b+1]^2=a^2frac(1)(16)+b^2frac(1)(4)$

$+1+abfrac(1)(4)-b+afrac(1)(4)$

Answer :

$=a^2frac(1)(16)+b^2frac(1)(4)+1+abfrac(1)(4)-b+afrac(1)(4)$

Q5. Factorise :

i) $4x^2+9y^2+16z^2+12xy-24yz-16xz$

ii) $2x^2+y^2+8z^2-2sqrt(2)xy+4sqrt(2)yz-8xz$

Sol. :

$4x^2+9y^2+16z^2+12xy-24yz-16xz$

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

$+2(3y)(-4z)+2(-4z)(2x)$

$[p^2+q^2+r^2+2pq+2qr+2rp=(p+q+r)^2]$ [We use an identity]

Where $p=2x, q=3y$ and $r=-4z$

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

Answer :

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

ii)

$2x^2+y^2+8z^2-2sqrt(2)xy+4sqrt(2)yz-8xz$

$=(-sqrt(2)x)^2+(y)^2+(2sqrt(2)x)^2+2(-sqrt(2)x)(y)$

$+2(y)(2sqrt(2)z)+2(2sqrt(2)z)(-sqrt(2)x)$

$[p^2+q^2+r^2+2pq+2qr+2rp=(p+q+r)^2]$ [We use an identity]

Where $p=-sqrt(2)x, q=y$ and $r=2sqrt(2)z$

$=(-sqrt(2)x+y+2sqrt(2)z)^2$

Answer :

$=(-sqrt(2)x+y+2sqrt(2)z)^2$

Q6. Write the following cubes in expanded form :

i) $(2x+1)^3$ ii) $(2a-3b)^3$

iii) $[frac(3)(2)x+1]^5$ iv) $[x-frac(2)(3)y]^3$

Sol. :

$(2x+1)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=2x$ , and

$q=1$

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

$=8x^3+12x^2+6x+1$

Answer :

$=8x^3+12x^2+6x+1$

ii)

$(2a-3b)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=2a$ and

$q=-3b$

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

$+(-3b)^3$

$=8a^3-36a^2b+54ab^2-27b^3$

Answer :

$=8a^3-36a^2b+54ab^2-27b^3$

iii)

$(frac(3)(2)x+1)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=frac(3)(2)x$ , and

$q=1$

$=(frac(3)(2)x)^3+3(frac(3)(2)x)^21+3(frac(3)(2)x)(1)^2$

$+(1)^3$

$=frac(27)(8)x^3+3frac(9)(4)x^2+3frac(3)(2)x+1$

$=frac(27)(8)x^3+frac(27)(4)x^2+frac(9)(2)+1$

Answer :

$=frac(27)(8)x^3+frac(27)(4)x^2+frac(9)(2)+1$

iv)

$(x-frac(2)(3)y)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=x$ and

$q=-frac(2)(3)y$

$=(x)^3+3(x)^2(-frac(2)(3)y)+3x(-frac(2)(3)y)^2$

$+(-frac(2)(3)y)^3$

$=x^3-2x^2y+frac(4)(3)xy^2-frac(8)(27)y^3$

Answer :

$=x^3-2x^2y+frac(4)(3)xy^2-frac(8)(27)y^3$

Q7. Evaluate the following using suitable identities :

i) $(99)^3$ ii) $(102)^3$ iii) $(998)^3$

Sol. :

$(99)^3$

$=(100-1)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=100$ , and

$q=-1$

$=(100)^3+3(100)^2(-1)+3(100)(-1)^2+ (1)^3$

$=1000000-30000+300+1$

$=970299$

Answer :

$=970299$

ii)

$(102)^3$

$=(100+2)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=100$ and

$q=2$

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

$=1000000+60000+1200+8$

$=1061208$

Answer :

$=1061208$

iii)

$(998)^3$

$=(1000-2)^3$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

Where

$p=1000$ , and

$q=-2$

$=(1000)^3+3(1000)^2(-2)+3(1000)(-2)^2+(-2)^3$

$=1000000000-6000000+6000-8$

$=994011992$

Answer :

$=994011992$

Q8. Factorise each of the following :

i) $8a^3+b^3+12a^2b+6ab^2$

ii) $8a^3-b^3-12a^2b+6ab^2$

iii) $27-125a^3-135a+225a^2$

iv) $64a^3-27b^3-144a^2b+108ab^2$

Sol. :

$8a^3+b^3+12a^2b+6ab^2$

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

$p^3+q^3+3p^2q+3pq^2=(p+q)^3$ [We use an identity]

Where

$p=2a$ , and

$q=b$

$=(2a+b)^3$

Answer :

$=(2a+b)^3$

ii)

$8a^3-b^3-12a^2b+6ab^2$

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

$p^3-q^3-3p^2q+3pq^2=(p-q)^3$ [We use an identity]

Where

$p=2a$ , and

$q=-b$

$=(2a-b)^3$

Answer :

$=(2a-b)^3$

iii)

$27-125a^3-135a+225a^2$

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

$p^3-q^3-3p^2q+3pq^2=(p-q)^3$ [We use an identity]

Where

$p=3$ , and

$q=5a$

$=(3-5a)^3$

Answer :

$=(3-5a)^3$

iv)

$64a^3-27b^3-144a^2b+108ab^2$

$=(4a)^3-(3b)^3-3(4a)^2(3b)+3(4a)(3b)^2$

$p^3-q^3-3p^2q+3pq^2=(p-q)^3$ [We use an identity]

$=(4a-3b)^3$

Answer :

$=(4a-3b)^3$

Q9. Verify :

(i) $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

ii) $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Sol. :

(i)

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

$(p+q)^3=p^3+3p^2q+3pq^2+q^3$ [We use an identity]

$p^3+q^3=(p+q)^3-3p^2q-3pq^2$

$p^3+q^3=(p+q)^3-3pq(p+q)$

$p^3+q^3=(p+q)[(p+q)^2-3pq]$

$p^3+q^3=(p+q)[(p^2+q^2+2pq-3pq]$

$p^3+q^3=(p+q)(p^2-pq+q^2)$

Where

$p=x$ , and

$q=y$

$x^3+y^3=(x+y)(x^2-xy+y^2)$ .

Proved

Alternate method

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

RHS

$=(x+y)(x^2-xy+y^2)$

$=x(x^2-xy+y^2)+y(x^2-xy+y^2)$

$=x^3-x^2y+xy^2+x^2y-xy^2+y^3$

$=x^3+y^3$

=LHS.

Proved.

ii)

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

RHS

$=(x-y)(x^2+xy+y^2)$

$=x(x^2+xy+y^2)-y(x^2+xy+y^2)$

$=x^3+x^2y+xy^2-x^2y-xy^2-y^3$

$=x^3-y^3$

=LHS

proved.

Q10. Factorise each of the following :

i) $27y^3 + 125z^3$ ii) $64m^3 - 343n^3$

Sol. :

$27y^3+125z^3$

$=(3y)^3+(5z)^3$

$[p^3+q^3=(p+q)(p^2-pq+q^2]$ [We use an identity]

Where

$p=3y$ , and

$q=5z$

$=(3y+5z)^3[(3y)^2-(3y)(5z)+(5z)^2]$

$=(3y+5z)^3(9y^2-15yz+25z^2)$

Answer :

$=(3y+5z)(9y^2-15yz+25z^2)$

ii)

$64m^3 - 343n^3$

$=(4m)^3-(7n)^3$

$[p^3-q^3=(p-q)(p^2+pq+q^2]$ [We use an identity]

Where

$p=4m$ , and

$q=7n$

$=(4m-7n)[(4m)^2+(4m)(7n)+(7n)^2]$

$=(4m-7n)(16m^2+28mn+49n^2)$

Answer :

$=(4m-7n)(16m^2+28mn+49n^2)$

Q11. Factorise $27x^3 + y^3 + z^3 - 9xyz$

Sol. :

$27x^3 + y^3 + z^3 - 9xyz$

$=(3x)^3+(y)^3+(z)^3-3(3x)(y)(z)$

$p^3+q^3+r^3-3pqr=(p+q+r)[p^2+q^2+r^2-pq-qr-rp]$ [We use an identity]

Where

$p=3x, q=y$ , and

$r=z$

$=(3x+y+z)[(3x)^2+(y)^2+(z)^2-3xy$

`-yz-3xz]`

$=(3x+y+z)[9x^2+y^2+z^2-3xy$

$-yz-3xz]$

Answer :

$=(3x+y+z)[9x^2+y^2+z^2-3xy$

$-yz-3xz]$

Q12. Verify that $x^3 + y^3 + z^3 - 3xyz$

$= frac(1)(2)(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$

Sol. :

RHS.

$=frac(1)(2)(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$

$=frac(1)(2)(x+y+z)[x^2+y^2-2xy+y^2+z^2-2yz$

$+z^2+x^2-2xz]$

$=frac(1)(2)(x+y+z)[2x^2+2y^2+2z^2-2xy$

$-2yz-2xz]$

$=frac(1)(2)(x+y+z)2×[x^2+y^2+z^2-xy-yz-xz]$

$=(x+y+z)[x^2+y^2+z^2-xy-yz-xz]$

$(p+q+r)[p^2+q^2+r^2-pq-qr-rp]$

$=p^3+q^3+r^3-3pqr$ [We use an identity]

Where

$p=x, q=y$ , and

$r=z$

$=x^3+y^3+z^3-3xyz$

=LHS.

Proved.

Q13. If $x + y + z = 0$ , show that $x^3 + y^3 + z^3 = 3xyz$ .

Sol. :

We know that

$p^3+q^3+r^3-3pqr=(p+q+r)[p^2+q^2+r^2-pq-qr-rp]$ [We use an identity]

Here $x+y+z=0$ [Given]

$x^3+y^3+z^3-3xyz=0×(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz=0$

$x^3+y^3+z^3=3xyz$

Proved

Q14.Without actually calculating the cubes, find the value of each of the following :

i) $(-12)^3 + (7)^3 + (5)^3$

ii) $(28)^3 + (-15)^3 + (-13)^3$

Sol. :

$(-12)^3 + (7)^3 + (5)^3$

$p^3+q^3+r^3-3pqr=(p+q+r)[p^2+q^2+r^2-pq-qr-rp]$ [We use an identity]

Where

$p=-12, q=7,$ and

$r=5$

$(-12)^3+(7)^3+(5)^3-3(-12)(7)(5)=(-12+7+5)[(-12)^2+(7)^2+(5)^2$

$-(-12)(7)-(7)(5)-(5)(-12)]$

$(-12)^3+(7)^3+(5)^3-3(-12)(7)(5)=0$

$(-12)^3+(7)^3+(5)^3=3(-12)(7)(5)$

$(-12)^3+(7)^3+(5)^3=-1260$

Answer :

$=-1260$

ii)

$(28)^3 + (-15)^3 + (-13)^3$

$p^3+q^3+r^3-3pqr=(p+q+r)[p^2+q^2+r^2-pq-qr-rp]$ [We use an identity]

Where

$p=28, q=-15,$ and

$r=-13$

$(28)^3+(-15)^3+(-13)^3-3(28)(-15)(-13)=(28-15-13)[(28)^2+(-15)^2$

$+(-13)^2-(28)(-15)-(-15)(-13)-(-13)(28)]$

$(28)^3+(-15)^3+(-13)^3-3(28)(-15)(-13)=0$

$(28)^3+(-15)^3+(-13)^3=3(28)(-15)(-13)$

$(28)^3+(-15)^3+(-13)^3=-16380$

Answer :

$=16380$

Q15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :

i) Area : $25a^2 - 35a + 12$ ii) Area : $35y^2 + 13y - 12$

Sol. :

$Area=25a^2 - 35a + 12$

$Area=25a^2-20a-15a+12$

$Area=5a(5a-4)-3(5a-4)$

$Area=(5a-4)(5a-3)$

$Area=L×B$

Answer :

$L=(5a-4)$ , and

$B=(5a-3)$

ii)

$Area=35y^2 + 13y - 12$

$Area=35y^2+28y-15y-12$

$Area=7y(5y+4)-3(5y+4)$

$Area=(5y+4)(7y-3)$

$Area=L×B$

Answer :

$L=(5y+4)$ , and

$B=(7y-3)$

Q16. What are the possible expressions for the dimensions of the cuboids, whose volumes are given below?

i) Volume : $3x^3-12x$ ii) Volume : $12ky^2+8ky-20k$

Sol. :

$Volume=3x^3-12x$

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

$Volume=3x[(x)^2-(2)^2]$

$Volume=3x(x+2)(x-2)$

$Volume=L×B×H$

Answer :

$L=3x, B=(x+2),$ and

$H=(x-2)$

ii)

$Volume=12ky^2+8ky-20k$

$Volume=4k[3y^2+2y-5]$

$Volume=4k[3y^2+5y-3y-5]$

$Volume=4k[y(3y+5)-1(3y-5)]$

$Volume=4k(3y+5)(y-1)$

$Volume=L×B×H$

Answer :

$L=4k, B=(3y+5),$ and

$H=(y-1)$

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9th Maths 2.5

NCERT Class 9th solution of Exercise 2.1

NCERT Class 9th solution of Exercise 2.2

NCERT Class 9th solution of Exercise 2.3

NCERT Class 9th solution of Exercise 2.4

NCERT Class 9th Maths Projects

Exercise 2.5

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