Question 12 Marks
If $x + 5y = 10$; find the value of $x^3+ 125y^3+ 150xy - 1000.$
Answer$x + 5y = 10$
$\Rightarrow (x + 5y)^3= 10^3$
$\Rightarrow x^3+ (5y)^3+ 3(x)(5y)(x + 5y) = 1000$
$\Rightarrow x^3+ (5y)^3+ 3(x)(5y)(10) = 1000$
$= x^3+ (5y)^3+ 150xy = 1000$
$= x^3+ (5y)^3 + 150xy - 1000$
$= 0$
View full question & answer→Question 22 Marks
If $a - 2b + 3c = 0;$ state the value of $a^3- 8b^3+ 27c^3.$
Answer$a^3- 8b^3+ 27c^3= a^3+ (-2b)^3+ (3c)^3$ Since $a - 2b + 3c = 0,$
we have
$a^3- 8b^3+ 27c^3= a^3+ (-2b)^3+ (3c)^3$
$= 3(a)( -2b)(3c)$
$= -18abc$
View full question & answer→Question 32 Marks
Using suitable identity $,$ evaluate $(97)^3$
Answer$(97)^3= (100 - 3)^3$
$= (100)^3- (3)^3- 3 \times 100 \times 3(100 - 3)$
$= 1000000 - 27 - 900 \times 97$
$= 1000000 - 27 - 87300$
$= 912673$
View full question & answer→Question 42 Marks
Using suitable identity $,$ evaluate $(104)^3$
AnswerUsing identity$: (a ± b)^3= a^3± b^3± 3ab(a ± b)$
$(104)^3= (100 + 4)^3$
$= (100)^3+ (4)^3+ 3 \times 100 \times 4(100 + 4)$
$= 1000000 + 64 + 1200 \times 104$
$= 1000000 + 64 + 124800$
$= 1124864$
View full question & answer→Question 52 Marks
Simplify using following identity :$(a \pm b)\left(a^2 \pm a b+b^2\right)=a^3 \pm b^3 \left(\frac{a}{3}-3 b\right)\left(\frac{a^2}{9}+a b+9 b^2\right)$
Answer$ \left(\frac{a}{3}-3 b\right)\left(\frac{a^2}{9}+a b+9 b^2\right)$
$ =\left(\frac{a}{3}-3 b\right)\left[\left(\frac{a}{3}\right)^2+\left(\frac{a}{3}\right)(3 b)+(3 b)^2\right]$
$ =\left(\frac{a}{3}\right)^3-(3 b)^3$
$ =\frac{a^3}{27}-27 b^3$
View full question & answer→Question 62 Marks
Simplify using following identity
$(a \pm b)\left(a^2 \pm a b+b^2\right)=a^3 \pm b^3$
$ \left(3 x-\frac{5}{x}\right)\left(9 x^2+15+\frac{25}{x^2}\right)$
Answer$ \left(3 x-\frac{5}{x}\right)\left(9 x^2+15+\frac{25}{x^2}\right)$
$ =\left(3 x-\frac{5}{x}\right)\left[(3 x)^2+(3 x)\left(\frac{5}{x}\right)+\left(\frac{5}{x}\right)^2\right]$
$ =(3 x)^3-\left(\frac{5}{x}\right)^3$
$ =27 x^3-\frac{125}{x^3}$
View full question & answer→Question 72 Marks
Simplify using following identity$ :(a \pm b)\left(a^2 \pm a b+b^2\right)=a^3 \pm b^3$
$ (2 x+3 y)\left(4 x^2+6 x y+9 y^2\right)$
Answer$( 2x + 3y )( 4x^2 + 6xy + 9y^2)$
$= ( 2x + 3y )[ (2x)^2 - (2x)(3y) + (3y)^2]$
$= (2x)^3 + (3y)^3$
$= 8x^3 + 27y^3$
View full question & answer→Question 82 Marks
Simplify : $( x - 6 )( x - 4 )( x - 2 )$
AnswerUsing identity :
$(x + a)(x + b)(x + c) = x^3+ (a + b + c)x^2+ (ab+bc+ ca)x +abc$
$( x - 6 )( x - 4 )( x - 2 )$
$= x^3+ (-6 - 4 - 2)x^2 + [-6 \times (-4) + (-4) \times (-2) + (-2) \times (-6)]x + (-6) \times (-4) \times (-2)$
$= x^3- 12x^2+ (24 + 8 + 12)x - 48$
$= x^3- 12x^2 + 44x - 48$
View full question & answer→Question 92 Marks
Simplify : $( x - 6 )( x - 4 )( x + 2 )$
AnswerUsing identity : $(x + a)(x + b)(x + c)$
$= x^3+ (a + b + c)x^2+ (ab+bc+ ca)x +abc$
$(x - 6)(x - 4)(x + 2)$
$= x^3+ (-6 - 4 + 2)x^2+ [-6 \times (-4) + (-4) \times 2 + 2 \times (-6)]x + (-6) \times (-4) \times 2$
$= x^3- 8x^2+ (24 - 8 - 12)x + 48$
$= x^3- 8x^2+ 4x + 48$
View full question & answer→Question 102 Marks
Simplify : $( x + 6 )( x + 4 )( x - 2 )$
AnswerUsing identity :
$(x + a)(x + b)(x + c)$
$= x^3+ (a + b + c)x^2+ (ab+bc+ ca)x +abc$
$(x + 6)(x + 4)(x - 2)$
$= x^3+ (6 + 4 - 2)x^2+ [6 \times 4 + 4 \times (-2) + (-2) \times 6]x + 6 \times 4 \times (-2)$
$= x^3+ 8x^2+ (24 - 8 - 12)x - 48$
$= x^3+ 8x^2+ 4x - 48$
View full question & answer→Question 112 Marks
Find : $(a + b)(a + b)$
Answer$(a + b)(a + b)$
$= (a + b)^2$
$= a \times a + a \times b + b \times a + b \times b$
$= a^2+ab+ab+ b^2$
$= a^2+ b^2+ 2ab$
View full question & answer→Question 122 Marks
If $a^2+ b^2= 34$ and $ab = 12;$ find$ : 7(a - b)^2- 2(a + b)^2$
Answer$a^2+b^2= 34,ab= 12$
$(a+b)^2=a^2+b^2+2ab$
$= 34 + 2 \times 12 = 34 + 24 = 58$
$(a-b)^2=a^2+b^2-2ab$
$= 34 - 2 x 12 = 34- 24 = 10$
$7(a-b)^2-2(a+b)^2= 7 \times 10 - 2 \times 58 = 70 - 116 = - 46$
View full question & answer→Question 132 Marks
If $a^2+ b^2 = 34$ and $ab = 12; $find $: 3(a + b)^2+ 5(a - b)^2$
Answer$a^2+b^2= 34,ab= 12$
$(a+b)^2=a^2+b^2+2ab$
$= 34 + 2 \times 12 = 34 + 24 = 58$
$(a-b)^2=a^2+b^2-2ab$
$= 34 - 2 \times 12 = 34- 24 = 10$
$3(a + b)^2+5(a-b)^2$
$= 3 \times 58 + 5 \times 10$
$= 174 + 50$
$= 224$
View full question & answer→Question 142 Marks
In the expansion of $(2x^2- 8) (x - 4)^2;$ find the value of constant term.
Answer$( 2x^2 - 8 )( x - 4 )^2$
$= ( 2x^2 - 8 )( x^2 - 8x + 16 )$
$= 4x^4 - 16x^3 + 32x^2 - 8x^2 + 64x -128$
$= 4x^4 - 16x^3 + 24x^2 + 64x - 128$
Hence,
Constant term $= -128$
View full question & answer→Question 152 Marks
In the expansion of $(2x^2- 8) (x - 4)^2;$ find the value of coefficient of $x^2$
Answer$( 2x^2 - 8 )( x - 4 )^2$
$= ( 2x^2 - 8 )( x^2 - 8x + 16 )$
$= 2x^4 - 16x^3 + 32x^2 - 8x^2 + 64x -128$
$= 2x^4 - 16x^3 + 24x^2 + 64x - 128$
Hence,
Coefficient of $x^2 = 24$
View full question & answer→Question 162 Marks
In the expansion of $(2x^2- 8) (x - 4)^2; $ find the value of coefficient of $x^3.$
Answer$( 2x^2 - 8 )( x - 4 )^2$
$= ( 2x^2 - 8 )( x^2 - 8x + 16 )$
$= 2x^2( x^2 - 8x + 16 ) - 8( x^2 - 8x + 16 )$
$= 2x^4 - 16x^3 + 32x^2 - 8x^2 + 64x -128$
$= 2x^4 - 16x^3 + 24x^2 + 64x - 128$
Hence,
Coefficient of $x^3 = - 16$
View full question & answer→Question 172 Marks
If $a+\frac{1}{a}=m$ and $a \neq 0;$ find in terms of $' m\ ';$ the value of $: a^2-\frac{1}{a^2}$
Answer$a^2-\frac{1}{a^2}$
$=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)$
$\left[\right.$ Since $\left.a^2-b^2=(a+b)(a-b)\right]$
$ =m\left( \pm \sqrt{m^2-4}\right)$
$ = \pm m \sqrt{m^2-4}$
View full question & answer→Question 182 Marks
Find the value of $'a\ ': 9x^2+ (7a - 5)x + 25 = (3x + 5)^2$
Answer$9x^2+ (7a - 5)x + 25 = (3x + 5)^2$
Comparing coefficients of $x$ terms, we get
$(7a - 5)x = 30x$
$7a - 5 = 30$
$7a = 35$
$a = 5$
View full question & answer→Question 192 Marks
Find the value of $'a\ ':4x^2+ ax + 9 = (2x - 3)^2$
Answer$4x^2+ ax + 9 = (2x - 3)^2$
Comparing coefficients of $x$ terms, we get
$ax = -12x$
so$, a = -12$
View full question & answer→Question 202 Marks
Find the value of $'a\ ': 4x^2+ ax + 9 = (2x + 3)^2$
Answer$4x^2+ ax + 9 = (2x + 3)^2$
Comparing coefficients of $x$ terms, we get
$ax = 12x$
so$, a = 12$
View full question & answer→Question 212 Marks
The sum of two numbers is $7$ and the sum of their cubes is $133,$ find the sum of their square.
AnswerLet $a, b$ be the two numbers.
$.'.a+b= 7$ and $a^3+b^3= 133$a
$(a+b)^3=a^3+b^3+3ab (a+b)$
$\Rightarrow (7)^3= 133+3ab (7)$
$\Rightarrow 343 = 133 + 21ab$
$\Rightarrow 21ab = 343 - 133 = 210$
$\Rightarrow 21ab = 210$
$\Rightarrow ab= 10$
Nowa $2+b^2=(a+b)^2-2ab$
$= 7^2- 2 x 10$
$=49 - 20$
$=29$
View full question & answer→Question 222 Marks
If $3a + 5b + 4c = 0,$ show that $: 27a^3+ 125b^3+ 64c^3= 180abc$
AnswerGiven that $3a+5b+4c = 0$
$3a + 5b = - 4c$
Cubing both sides,
$(3a+5b)^3=(-4c)^3$
$\Rightarrow (3a)^3+(5b)^3+3\times 3a\times 5b (3a+5b) = -64c^3$
$\Rightarrow 27a^3+125b^3+45ab\times (-4c)=-64c^3$
$\Rightarrow 27a^3+125b^3-180abc = -64c^3$
$\Rightarrow 27a^3+125b^3+64c^3$
$= 180abc$
Hence proved.
View full question & answer→Question 232 Marks
If $x+ y - z = 4$ and $x^2+ y^2+ z^2= 30,$ then find the value of $xy-yz-zx.$
Answer$x+y-z=4$ and $x^2+y^2+z^2$
$=30$
Since $(x+y-z) 2=x 2+y 2+z 2+2(x y-y z-z x)$,
we have
$(4) 2=30+2(x y-y z-z x)$
$ \Rightarrow 16=30+2(x y-y z-z x)$
$ \Rightarrow 2(x y-y z-z x)=-14$
$ \Rightarrow x y-y z-z x=-\frac{14}{2}=-7$
$ \therefore x y-y z-z x=-7$
View full question & answer→Question 242 Marks
If $a^2+ b^2+ c^2= 50$ and $ab+bc+ ca = 47,$find $a + b + c.$
Answer$a^2+b^2+c^2=50$ and $a b+b c+c a=47 $
$ \text { Since }(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a) $
$ \therefore(a+b+c)^2=50+2(47) $
$ \Rightarrow(a+b+c)^2=50+94=144 $
$ \Rightarrow a+b+c=\sqrt{144}= \pm 12 $
$ \therefore a+b+c= \pm 12$
View full question & answer→Question 252 Marks
If $a + b + c = p$ and $ab + bc + ca = q ;$ find $a^2+ b^2+ c^2.$
AnswerWe know that
$( a + b + c )^2 = a^2 + b^2 + c^2 + 2( ab + bc + ca ) .....(1)$
Given that, $a + b + c = p$ and $ab + bc + ca = q$
We need to find $a^2 + b^2 + c^2 :$
Substitute the values of $( ab + bc + ca )$ and $( a + b + c )$
in the identity $(1),$ we have
$(p)^2 = a^2 + b^2 + c^2 + 2q$
$\Rightarrow p^2 = a^2 + b^2 + c^2 + 2q$
$\Rightarrow a^2 + b^2 + c^2$
$= p^2 - 2q$
View full question & answer→Question 262 Marks
Expand $: \left(x-\frac{1}{x}+5\right)^2$
Answer$ \left(x-\frac{1}{x}+5\right)^2$
$=(x)^2+\left(\frac{1}{x}\right)^2+(5)^2-2(x)\left(\frac{1}{x}\right)-2\left(\frac{1}{x}\right)(5)+2(5)(x)$
$ =x^2+\frac{1}{x^2}+25-2-\frac{10}{x}+10 x$
$ =x^2+\frac{1}{x^2}+23-\frac{10}{x}+10 x$
View full question & answer→Question 272 Marks
Expand$: \left(3 a+\frac{2}{b}\right)\left(2 a-\frac{3}{b}\right)$
Answer$ \left(3 a+\frac{2}{b}\right)\left(2 a-\frac{3}{b}\right)$
$=(3 a)(2 a)+\left(\frac{2}{b}\right)(2 a)-\left(\frac{3}{b}\right)(3 a)-\left(\frac{2}{b}\right)\left(\frac{3}{b}\right)$
$ =6 a^2+\left(\frac{4}{b}-\frac{9}{b}\right) a-\frac{6}{b^2}$
$ =6 a^2+\left(-\frac{5}{b}\right) a-\frac{6}{b^2}$
$ =6 a^2-\frac{5 a}{b}-\frac{6}{b^2}$
View full question & answer→Question 282 Marks
Expand $: \left(2 x-\frac{1}{x}\right)\left(3 x+\frac{2}{x}\right)$
Answer$ \left(2 x-\frac{1}{x}\right)\left(3 x+\frac{2}{x}\right)$
$ =2 x\left(3 x+\frac{2}{x}\right)-\frac{1}{x}\left(3 x+\frac{2}{x}\right)$
$ =6 x^2+2 x \times \frac{2}{x}-\frac{1}{x} \times 3 x-\frac{1}{x} \times \frac{2}{x}$
$ =6 x^2+4-3-\frac{2}{x^2}$
$ =6 x^2+1-\frac{2}{x^2}$
View full question & answer→Question 292 Marks
Expand $ : ( x - 8 )( x - 10 )$
Answer$( x - 8 )( x - 10 )$
$= x(x - 10) - 8(x - 10)$
$= x^2 - 10x - 8x + 80$
$= x^2 - 18x + 80$
View full question & answer→Question 302 Marks
If $a \neq 0$ and $a-\frac{1}{a}=3; $ find $:a^2+\frac{1}{a^2}$
Answer$a-\frac{1}{a}=3$
$\left(a-\frac{1}{a}\right)^2=9$
$a^2+\frac{1}{a^2}=9+2$
$=11$
View full question & answer→Question 312 Marks
Use property to evaluate : $38^3+ (-26)^3+ (-12)^3$
AnswerProperty is if $a + b + c = 0$ then $a^3+ b^3+ c^3= 3abc$
$a = 38, b = -26, c = -12$
$38^3+ (-26)^3+ (-12)^3$
$= 3(38)(-26)(-12)$
$= 35568$
View full question & answer→Question 322 Marks
Use property to evaluate : $9^3- 5^3- 4^3$
AnswerProperty is if $a + b + c = 0$ then $a^3+ b^3+ c^3= 3abc$
$a = 9, b = -5, c = -4$
$9^3- 5^3- 4^3$
$= 9^3+ (-5)^3+ (-4)^3$
$= 3(9)(-5)(-4)$
$= 540$
View full question & answer→Question 332 Marks
Use property to evaluate : $7^3+ 3^3+ (-10)^3$
AnswerProperty is if $a + b + c = 0$ then $a^3+ b^3+ c^3= 3abc$
$a = 7, b = 3, c = -10$
$7^3+ 3^3+ (-10)^3$
$= 3(7)(3)(-10) $
$= -630$
View full question & answer→Question 342 Marks
Use property to evaluate : $13^3+ (-8)^3+ (-5)^3$
AnswerProperty is if $a + b + c = 0$ then $a^3+ b^3+ c^3= 3abc$
$a = 13, b = -8$ and $c = -5$
$13^3+ (-8)^3+ (-5)^3$
$= 3(13)(-8)(-5)$
$= 1560$
View full question & answer→Question 352 Marks
If $a + 2b = 5;$ then show that $: a^3+ 8b^3+ 30ab = 125.$
AnswerGiven that $a + 2b = 5$
We need to find $a^3 + 8b^3 + 30ab$
Now consider the cube of $a + 2b$
$( a + 2b )^3 = a^3 + (2b)^3 + 3 \times a \times 2b \times ( a + 2b )$
$( a + 2b )^3 = a^3 + 8b^3 + 6ab \times ( a + 2b )$
$5^3 = a^3 + 8b^3 + 6ab \times 5 [ \because a + 2b = 5 ]$
$125 = a^3 + 8b^3 + 30ab$
Thus the value of $a^3 + 8b^3 + 30ab $ is $125.$
View full question & answer→Question 362 Marks
If $a+\frac{1}{a}=p$ and $a \neq 0; $ then show that $: a^3+\frac{1}{a^3}=p\left(p^2-3\right)$
AnswerGiven that $a+\frac{1}{a}=\mathrm{p} .....(1)$
$ \left(a+\frac{1}{a}\right)^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)$
$ \Rightarrow a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)$
$ \Rightarrow a^3+\frac{1}{a^3}=(p)^3-3(p) \text { [ From(1)] }$
$\Rightarrow a^3+\frac{1}{a^3}$
$=p\left(p^2-3\right)$
View full question & answer→Question 372 Marks
If $4x^2+ y^2= a$ and $xy= b$, find the value of $2x + y.$
Answer$x y=a b ....(i)$
$4 x^2+y^2=a ....(ii)$
Now, $(2 x+y)^2=(2 x)^2+4 x y+y^2$
$=4 x^2+y^2+4 x y$
$=a+4 b$
$....[$From $(i)$ and $(ii)]$
$\Rightarrow 2 x+y= \pm \sqrt{a+4 b}$
View full question & answer→Question 382 Marks
Find the cube of$ : \left(3 a-\frac{1}{a}\right)(a \neq 0)$
Answer$ (a-b)^3=a^3-3 a^2 b+3 a b^2-b^3$
$ \left(3 a-\frac{1}{a}\right)^3$
$ =(3 a)^3-3 \times(3 a)^2 \times \frac{1}{a}+3 \cdot 3 a\left(\frac{1}{a}\right)^2-\left(\frac{1}{a}\right)^3$
$ =27 a^3-3 \cdot 9 a^2 \cdot \frac{1}{a}+9 a \cdot \frac{1}{a^2}-\frac{1}{a^3}$
$ =27 a^3-27 a+\frac{9}{a}-\frac{1}{a^3} .$
View full question & answer→Question 392 Marks
Find the cube of $: 2 a+\frac{1}{2 a} \quad(a \neq 0)$
Answer$(a+b) 3=a 3+3 a b(a+b)+b 3$
$=(2 a)^3+\left(\frac{1}{2 a}\right)^3$
$=3 \times 2 a \times \frac{1}{2 a}\left(2 a+\frac{1}{2 a}\right)$
$=8 a^3+\frac{1}{8 a^3}+3\left(2 a+\frac{1}{2 a}\right)$
$=8 a^3+\frac{1}{8 a^3}+6 a+\frac{3}{2 a}$
View full question & answer→Question 402 Marks
Expand : $(3x - 5y - 2z) (3x - 5y + 2z)$
Answer$(3x-5y-2z) (3x-5y+2z)$
$= {(3x-5y)-(2z)} {(3x-5y)+(2z)}$
$= (3x-5y)^2-(2z)^2{since(a+b) (a-b)=a^2-b^2}$
$= 9x^2+25y^2- 2 \times 3x\times 5y-4z^2$
$= 9x^2+25y^2-30xy-4z^2$
$= 9x^2+25y^2-4z^2-30xy$
View full question & answer→Question 412 Marks
Expand :$ (3x + 5y + 2z) (3x - 5y + 2z)$
Answer$(3x+5y+2z) (3x-5y+2z)$
$= {(3x+2z)+(5y)} {(3x+2z)-(5y)}$
$= (3x+2z)^2-(5y)^2$
${since(a+b) (a-b)=a^2-b^2}$
$= 9x^2+4z^2+2 \times 3x\times 2z -25y^2$
$= 9x^2+4z^2+12xz- 25y^2$
$= 9x^2+4z^2-25y^2+12xz$
View full question & answer→Question 422 Marks
If $2x - 3y = 10$ and $xy= 16;$ find the value of $8x^3- 27y^3.$
AnswerGiven that $2x-3y= 10,xy= 16$
$\therefore (2x-3y)^3=(10)^3$
$\Rightarrow 8x^3-27y^3- 3(2x) (3y) (2x-3y) = 1000$
$\Rightarrow 8x^3- 27y^3-18xy (2x-3y) = 1000$
$\Rightarrow 8x^3-27 y^3- 18 (16) (10)=1000$
$\Rightarrow 8x^3-27 y^3- 2880=1000$
$\Rightarrow 8x^3-27 y^3=1000+2880$
$\Rightarrow 8x^3-27 y^3$
$=3880$
View full question & answer→Question 432 Marks
Find the cube of : $5a + 3b$
Answer$( a + b )^3 = a^3 + 3ab( a + b ) + b^3$
$( 5a + 3b)^3 = (5a)^3 + 3 \times 5a \times 3b( 5a + 3b) + (3b)^3$
$= 125a^3 + 45ab( 5a + 3b ) + 27b^3$
$= 125a^3 + 225a^2b + 135ab^2 + 27b^3$
View full question & answer→Question 442 Marks
Find the cube of : $3a- 2b$
Answer$( a - b )^3 = a^3 - 3ab( a - b ) - b^3$
$( 3a - 2b )^3 = (3a)^3 - 3 \times 3a \times 2b( 3a - 2b) - (2b)^3$
$= 27a^3 - 18ab( 3a - 2b ) - 8b^3$
$= 27a^3 - 54a^2b + 36ab^2 - 8b^3$
View full question & answer→Question 452 Marks
If $a \neq 0$ and $a-\frac{1}{a}=4;$ find $: \left(a^3-\frac{1}{a^3}\right)$
Answer$ \left(a-\frac{1}{a}\right)^3=a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)$
$ \Rightarrow\left(a^3-\frac{1}{a^3}\right)=\left(a-\frac{1}{a}\right)^3+3\left(a-\frac{1}{a}\right)$
$ \Rightarrow\left(a^3-\frac{1}{a^3}\right)=(4)^3+3(4) \quad\left[\because a-\frac{1}{a}=4\right]$
$ \Rightarrow\left(a^3-\frac{1}{a^3}\right)=64+12$
$ \Rightarrow\left(a^3-\frac{1}{a^3}\right)$
$=76$
View full question & answer→Question 462 Marks
If $\mathrm{a} \neq 0$ and $a-\frac{1}{a}=4 ;$ find $: \left(a^2+\frac{1}{a^2}\right)$
Answer$ \left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$ \Rightarrow a^2+\frac{1}{a^2}=\left(a-\frac{1}{a}\right)^2+2$
$ \Rightarrow a^2+\frac{1}{a^2}=(4)^2+2 \quad\left[\because a-\frac{1}{a}=4\right]$
$ \Rightarrow a^2+\frac{1}{a^2}$
$=18$
View full question & answer→Question 472 Marks
Evaluate : $(4a +3b)^2- (4a - 3b)^2+ 48ab.$
AnswerBy using for this formula $=x^2-y^2=(x+y)(x-y)$
$=(4 a+3 b+4 a-3 b)(4 a+3 b-4 a+3 b)+48 a b$
$=(8 a)(6 b)+48 a b$
$=48 a b+48 a b$
$=96 a b$
View full question & answer→Question 482 Marks
Evalute $: \left(\frac{2 x}{7}-\frac{7 y}{4}\right)^2$
Answer$\left(\frac{2 x}{7}-\frac{7 y}{4}\right)^2$
We know that
$(a-b)^2=a^2-2 a b+b^2$
$\therefore\left(\frac{2 x}{7}-\frac{7 y}{4}\right)^2$
$=\left(\frac{2 x}{7}\right)^2+\left(\frac{7 y}{4}\right)^2-2 \times \frac{2 x}{7} \times \frac{7 y}{4}$
$ =\frac{4 x^2}{49}+\frac{49 y^2}{16}-x y$
View full question & answer→Question 492 Marks
Use identities to evaluate : $(998)^2$
Answer$(998)^2$
$(998)^2 = (1000 - 2)^2$
We know that
$( a - b )^2 = a^2 + b^2 - 2ab$
$\therefore (1000 - 2)^2$
$= 1000^2 + 2^2 - 2 \times 1000 \times 2$
$= 1000000 + 4 - 4000$
$= 9,96,004$
View full question & answer→Question 502 Marks
Use identities to evaluate : $(97)^2$
Answer$(97)^2$
$(97)^2 = (100 - 3)^2$
We know that,
$( a - b )^2 = a^2 + b^2 - 2ab$
$\therefore (100 - 3)^2$
$= 100^2 + 3^2 - 2 \times 100\times 3$
$= 10000 + 9 - 600$
$= 9,409$
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