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MATHEMATICS [4 marks sum]

Expansions (Including Substitution) · ICSE STD 9 (ICSE - English Medium). 17 questions.

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Question 14 Marks
Simplify :$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
Answer
$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
If $a+b+c=0$,
then $a^3+b^3+c^3=3 a b c$
Now,
$\mathrm{x}^2-\mathrm{y}^2+\mathrm{y}^2-\mathrm{z}^2+\mathrm{z}^2-\mathrm{x}^2=0$
$\Rightarrow\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3$
$=3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)\dots..(1)$
And,
$x-y+y-z+z-x=0$
$\Rightarrow(x-y)^3+(y-z)^3+(z-x)^3$
$=3(x-y)(y-z)(z-x)\dots ....(2)$
Now,
$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
$ =\frac{3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)}{3(x-y)(y-z)(z-x)}$
$ =(x+y)(y+z)(z+x)\dots...[$From $(1)$ and $(2)]$
 
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Question 24 Marks
The difference between two positive numbers is $4$ and the difference between their cubes is $316.$Find : The sum of their squares
Answer
Given difference between two positive numbers is $4$ and difference between their cubes is $316.$
Let the positive numbers be $a$ and $b$
$a - b = 4 .....(1)$
$a^3- b^3= 316 .....(2)$
$ab = 21 ......(3)$
Squaring $(1)$ both sides,
 we get
$(a - b)^2= 16$
$a^2+ b^2- 2ab = 16$
$a^2+b^2=2 \times 21+16$
$a^2+b^2=42+16$
$a^2+b^2=58$
Sum of their squares is $58.$
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Question 34 Marks
If $\frac{x^2+1}{x}=3 \frac{1}{3}$ and $\mathrm{x}>1 ;$ Find If $x^3-\frac{1}{x^3}$
Answer
Given $\frac{x^2+1}{x}=3 \frac{1}{3}$
$\frac{x^2+1}{x}=\frac{10}{3}$
$ {\left[x+\frac{1}{x}\right]=\frac{10}{3}}$
Squaring on both sides,we get
$x^2+\frac{1}{x^2}+2=\frac{100}{9}$
$ x^2+\frac{1}{x^2}=\frac{100-18}{9}=\frac{82}{9}$
$ x-\frac{1}{x}=\sqrt{\left(x+\frac{1}{x}\right)^2-4}$
$=\sqrt{\frac{100}{9}-4}=\sqrt{\frac{64}{9}}=\frac{8}{3}$
$ \therefore x-\frac{1}{x}=\frac{8}{3}$
Cubing both sides, we get
$\left(x-\frac{1}{x}\right)^3=\frac{512}{27}$
$ x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=\frac{512}{27}$
$ x^3-\frac{1}{x^3}=\frac{512}{27}+8$
$=\frac{512+216}{27}$
$=\frac{728}{27}$
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Question 44 Marks
If $a^2+\frac{1}{a^2}=47$ and $a \neq 0$ find$:(i) a+\frac{1}{a},(ii)a^3+\frac{1}{a^3}$
Answer
$ \text { (i) } a+\frac{1}{a}$
$ a^2+\frac{1}{a^2}=47$
$ \left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=47+2$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=49$
$ \Rightarrow a+\frac{1}{a}= \pm \sqrt{49}$
$ \Rightarrow a+\frac{1}{a}= \pm 7$
$ \text { (ii) } a^3+\frac{1}{a^3}$
$ \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}=( \pm 7)^3-3( \pm 7) \quad[$ From $(1)]$
$ \Rightarrow a^3+\frac{1}{a^3}= \pm 322$
 
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Question 54 Marks
If $X \neq 0$ and $X+\frac{1}{X}=2$; then show that:$x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}$
Answer
$ \left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2$
$ \Rightarrow x^2+\frac{1}{x^2}=(2)^2-2 \quad\left[\because x+\frac{1}{x}=2\right]$
$\Rightarrow x^2+\frac{1}{x^2}=2 \dots(1)$
$ \left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)$
$ \Rightarrow x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$
$ \Rightarrow x^3+\frac{1}{x^3}=(2)^3-3(2) \quad\left[\because x+\frac{1}{x}=2\right]$
$\Rightarrow x^3+\frac{1}{x^3}=8-6$
$\Rightarrow x^3+\frac{1}{x^3}=2\ldots(2)$
We know that
$x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2$
$ =(2)^2-2 \quad$[ from$(1)]$
$ =4-2$
$\Rightarrow x^4+\frac{1}{x^4}=2\ldots(3)$
Thus from equations ( 1 ), ( 2 ) and (3), we have
$x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}$
$=x^4+\frac{1}{x^4}$
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Question 64 Marks
If $a^2-3 a+1=0$ and $a \neq 0$; find$:(i) a+\frac{1}{a},(ii) a^2+\frac{1}{a^2}$
Answer
$(i)$ Consider the given equation
$a^2-3 a+1=0$
Rewrite the given equation, we have
$a^2+1=3 a$
$ \Rightarrow \frac{a^2+1}{a}=3$
$ \Rightarrow\left[\frac{a^2}{a}+\frac{1}{a}\right]=3$
$\Rightarrow\left[a+\frac{1}{a}\right]=3\ldots(1)$
$(ii)$ We need to find $a^2+\frac{1}{a^2} :$
We know the identity,
$(a+b) ^2=a ^2+b ^2+2 a b$
$\therefore\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2\ldots . .(2)$
From equation $(1),$ we have,
$a+\frac{1}{a}=3$
Thus equation $(2),$ becomes,
$(3)^2=a^2+\frac{1}{a^2}+2$
$ \Rightarrow 9=a^2+\frac{1}{a^2}+2$
$ \Rightarrow a^2+\frac{1}{a^2}$
$=7$
 
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Question 74 Marks
If $x=3+2 \sqrt{ 2} $, find$(i) \frac{1}{x}(ii) x-\frac{1}{x}(iii) \left(x-\frac{1}{x}\right)^3(iv) x^3-\frac{1}{x^3}$
Answer
$ x=3+2 \sqrt{2}$
$ \text { (i) } \frac{1}{x}=\frac{1}{3+2 \sqrt{2}}$
$ =\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$
$ =\frac{3-2 \sqrt{2}}{(3)^2-(2 \sqrt{2})^2}$
$=\frac{3-2 \sqrt{2}}{9-8}$
$\therefore \frac{1}{x}=3-2 \sqrt{2}\ldots .(1)$
$(ii) x-\frac{1}{x}=(3+2 \sqrt{2})-(3-2 \sqrt{2}) \ldots[$ From$(2)]$
$=3+2 \sqrt{2}-3+2 \sqrt{2}$
$\therefore x-\frac{1}{x}=4 \sqrt{2}\ldots .(2)$
$(iii)\left(x-\frac{1}{x}\right)^3 =(4 \sqrt{2})^3$
$ =64 \times 2 \sqrt{2}$
$ =128 \sqrt{2}$
$(iv)x^3-\frac{1}{x^3}= \left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$ =128 \sqrt{2}+3(4 \sqrt{2})$
$ =128 \sqrt{2}+12 \sqrt{2}$
$ =140 \sqrt{2}$
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Question 84 Marks
If $x=\frac{1}{x-5}$ and $x \neq 5$. Find $: x^2-\frac{1}{x^2}$
Answer
Given $\mathrm{x}=\frac{1}{x-5}$;
By cross multiplication,
$\Rightarrow x(x-5)=1$
$ \Rightarrow x^2-5 x=1$
$\Rightarrow x^2-1=5 x\ldots .(1)$
Dividing both sides by $x$,
$\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2$
$ \Rightarrow(5)^2=x^2+\frac{1}{x^2}-2$
$\Rightarrow x^2+\frac{1}{x^2}=25+2=27\ldots(2)$
Let us consider the expansion of $\left(x+\frac{1}{x}\right)^2$ :
$\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(x+\frac{1}{x}\right)^2=27+2$
$ \Rightarrow\left(x+\frac{1}{x}\right)^2=29 [$from$(2)]$
$\Rightarrow\left(x+\frac{1}{x}\right)= \pm \sqrt{29}\ldots .(3)$
We know that,
$x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)=( \pm \sqrt{29})(5) [$From equation $(1)$ and $(3)]$
$x^2-\frac{1}{x^2}= \pm 5 \sqrt{29}$
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Question 94 Marks
If $x^2+x^{\frac{1}{2}}=7$ and $x \neq 0$; find the value of:$7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}$
Answer
Given that $x^2+\frac{1}{x^2}=7$
We need to find the value of $7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}$
Consider the given equation :
$x 2+\frac{1}{x^2}-2=7-2[$ subtract $2$ from both the sides $]$
$\Rightarrow\left(x-\frac{1}{x}\right)^2=5$
$\Rightarrow\left(x-\frac{1}{x}\right)= \pm \sqrt{5} \dots....(1)$
$\therefore 7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}=7 x^3-\frac{7}{x^3}+8 x-\frac{8}{x}$
$=7\left(x^3-\frac{1}{x^3}\right)+8\left(x-\frac{1}{x}\right)\ldots . . .(2)$
Now consider the expansion of $\left(x-\frac{1}{x}\right)^3$ :
$\left(x-\frac{1}{x}\right)^3=x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)$
$ \Rightarrow x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$\Rightarrow x^3-\frac{1}{x^3}=(\sqrt{5})^3+3(\sqrt{5})\ldots...(3)$.
Now substitute the value of $x^3-\frac{1}{x^3}$ in equation $(2)$, we have
$7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}=7\left(x^3-\frac{1}{x^3}\right)+8\left(x-\frac{1}{x}\right)$
$ \Rightarrow 7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}=7\left[(\sqrt{5})^3+3(\sqrt{5})\right]+8[\sqrt{5}][$ From$(3)]$
$ \left.\Rightarrow 7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}=7[5(\sqrt{5})+3(\sqrt{5})]+8[\sqrt{5}]\right]$
$ \Rightarrow 7 x^3+8 x-\frac{7}{x^3}-\frac{8}{x}=64[\sqrt{5}]$
 
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Question 104 Marks
If $2\left(x^2+1\right)=5 x$, find:$(i) x-\frac{1}{x};(ii) x^3-\frac{1}{x^3}$
Answer
$\text { (i) } 2\left(x^2+1\right)=5 x$
$ \left(x^2+1\right)=\frac{5}{2} x$
Dividing by $\mathrm{x}$, we have
$\frac{x^2+1}{x}=\frac{5}{2}$
$\Rightarrow\left(x+\frac{1}{x}\right)=\frac{5}{2}\ldots . .(1)$
Now consider the expansion of $\left(x+\frac{1}{x}\right)^2$ :
$\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2=x^2+\frac{1}{x^2}+2$
$ \Rightarrow\left(\frac{5}{2}\right)^2-2=x^2+\frac{1}{x^2}$
$ \Rightarrow \frac{25}{4}-2=x^2+\frac{1}{x^2}$
$ \Rightarrow x^2+\frac{1}{x^2}=\frac{25-8}{4}[$From$(1)]$
$\Rightarrow x^2+\frac{1}{x^2}=\frac{17}{4}\ldots .(2)$
Now consider the expansion of $\left(x-\frac{1}{x}\right)^2$ :
$\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17}{4}-2 [$from$(2)]$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{17-8}{4}$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=\frac{9}{4}$
$\Rightarrow\left(x-\frac{1}{x}\right)^2= \pm \frac{3}{2}\ldots .(3)$
$(ii)$ We know that,
$\left(x^3-\frac{1}{x^3}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$ \therefore\left(x^3-\frac{1}{x^3}\right)=\left( \pm \frac{3}{2}\right)^3+3\left( \pm \frac{3}{2}\right)[$from$(3)]$
$ = \pm \frac{27}{8}+\frac{9}{2}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{27+36}{8}$
$ \Rightarrow\left(x^3-\frac{1}{x^3}\right)= \pm \frac{63}{8}$

 
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Question 114 Marks
If $a - b = 4$ and $a + b = 6;$ find$(i) \ a^2+ b^2;(ii) \ ab$
Answer
$(i)$ We know that,
$(a-b)^2=a^2-2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a-b)+2 a b \dots....(1)$
Similarly, we know that, $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a+b)^2-2 a b\dots .....(2)$
Adding the equations $( 1 )$ and $( 2 ),$ we have
$2\left(a^2+b^2\right)=(a-b)^2+2 a b+(a+b)^2-2 a b$
$ \Rightarrow 2\left(a^2+b^2\right)=(a-b)^2+(a+b)^2$
$\Rightarrow\left(a^2+b^2\right)=\frac{1}{2}\left[(a-b)^2+(a+b)^2\right] \dots....(3)$
Given that $a+b=6 ; a-b=4$
Substitute the values of $(a+b)$ and $(a-b)$ in equation $(3),$ we have
$a^2+b^2=\frac{1}{2}\left[(4)^2+(6)^2\right]$
$ =\frac{1}{2}[16+36]$
$ =\frac{52}{2}$
$ \Rightarrow\left(a^2+b^2\right)=26\dots.....(4)$
From equation $(4),$ we have
$\mathrm{a} 2+\mathrm{b} 2=26$
Consider the identity,
$(a-b) 2=a 2+b 2-2 a b \dots....(5)$
Substitute the value $a-b=4$ and $a 2+b 2=26$
in the above equation, we have
$(4)\ 2=26-2 a b$
$ \Rightarrow 2 a b=26-16$
$ \Rightarrow 2 a b=10$
$ \Rightarrow a b=5$
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Question 124 Marks
If $a - b = 0.9$ and $ab = 0.36;$ find:$(i) \ a + b;(ii) \ a^2- b^2.$
Answer
$(i)$ We know that,
$(a-b)^2=a^2-2 a b+b^2$ and $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above equation, we have
$(a+b)^2 =a^2+b^2-2 a b+4 a b$
$ =(a-b)^2+4 a b$
Given that $a-b=0.9 ; a b=0.36$
Substitute the values of $(a-b)$ and $(ab)$
in equation $(1), $we have
$(a+b)^2 =(0.9)^2+4(0.36)$
$ =0.81+1.44=2.25$
$\Rightarrow a+b = \pm \sqrt{2.25}$
$\Rightarrow a+b = \pm 1.5$
$(ii)$ We know that,
$a^2-b^2=(a+b)(a-b)$
From equation $(2)$ we have,
$a+b= \pm 1.5$
Thus equation $(3)$ becomes,
$a^2-b^2=( \pm 1.5)(0.9) [$given $a-b=0.9]$
$\Rightarrow a^2-b^2= \pm 1.35 $
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Question 134 Marks
If $x+y=\frac{7}{2}$ and $x y=\frac{5}{2}$; find : $x-y$ and $x^2-y^2$
Answer
We know that,
$(x+y)^2=x^2+2 x y+y^2$
and
$(x-y)^2=x^2-2 x y+y^2$
Rewrite the above equation, we have
$(x-y)^2=x^2+y^2+2 x y-4 x y$
$=(x+y)^2-4 x y\ldots...(1)$
Given that $\mathrm{x}+\mathrm{y}=\frac{7}{2}$ and $\mathrm{xy}=\frac{5}{2}$ Substitute the values of $(x+y)$ and $(xy)$ in equation $(1)$, we have
$(x-y)^2=\left(\frac{7}{2}\right)^2-4\left(\frac{5}{2}\right)$
$ =\frac{49}{4}-10=\frac{9}{4}$
$ \Rightarrow x-y= \pm \sqrt{\frac{9}{4}}$
$\Rightarrow \mathrm{a}-\mathrm{b}= \pm\left(\frac{3}{2}\right)\dots...(2)$
We know that,
$x^2-y^2=(x+y)(x-y)\ldots...(3)$
From equation $(2)$ we have,
$x-y= \pm \frac{3}{2}$
Thus equation $(3)$ becomes,
$\mathrm{x}^2-\mathrm{y}^2=\left(\frac{7}{2}\right)\left( \pm \frac{3}{2}\right)[$given $\mathrm{x}+\mathrm{y}=\frac{7}{2}]$
$\Rightarrow \mathrm{x}^2-\mathrm{y}^2= \pm \frac{21}{4} $
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Question 144 Marks
Evaluate : $\left(\frac{a}{2 b}+\frac{2 b}{a}\right)^2-\left(\frac{a}{2 b}-\frac{2 b}{a}\right)^2-4$
Answer
Consider the given expression :
Let us expand the first term : $\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2$
We know that,
$(a+b)^2=a 2+b^2+2 a b$
$\therefore\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2+2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2\ldots(1)$
Let us expand the second term : $\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2$
We know that,
$(a-b)^2=a^2+b^2-2 a b$
$ \therefore\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2-2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}-2\ldots(2)$
Thus from $(1)$ and $(2)$, the given expression is
$\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2-\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2-4=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2-\frac{a^2}{(4 b)^2}-\frac{(4 b)^2}{a^2}+2-4$
$=0$.
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Question 154 Marks
If $a^2-5 a-1=0$ and $a \neq 0$; find:(i) $a-\frac{1}{a};(ii) a+\frac{1}{a};(iii);a^2-\frac{1}{a^2}$
Answer
$(i)$ Consider the given equation
$\text { a2 - } 5 \text { a - } 1=0$
Rewrite the given equation, we have
$\text { a2 - } 1=5 a$
$ \Rightarrow \frac{a^2-1}{a}=5$
$ \Rightarrow\left[\frac{a^2}{a}-\frac{1}{a}\right]=5$
$\Rightarrow a-\frac{1}{a}=5\ldots . .(1)$
$(ii)$ We need to find $a+\frac{1}{a}$ :
We know the identity, $(a-b)^2=a^2+b^2-2 a b$
$\therefore\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$\Rightarrow(5)^2=a^2+\frac{1}{a^2}-2 \quad[$ From $(1)]$
$\Rightarrow 25=a^2+\frac{1}{a^2}-2$
$\Rightarrow a^2+\frac{1}{a^2}=27\ldots .(2)$
Now consider the identity $(a+b)^2=a^2+b^2+2 a b$
$\therefore\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=27+2 \quad[$ From$(2)]$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=29$
$\Rightarrow a+\frac{1}{a}= \pm \sqrt{29} \dots.....(3)$
$(iii)$ We need to find $a^2-\frac{1}{a^2}$
We know the identity, $\mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$
$\therefore a^2-\frac{1}{a^2}=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right) \dots....(4)$
From equation $(3)$, we have,
$a+\frac{1}{a}= \pm \sqrt{29}$
From equation $(1)$, we have,
$a-\frac{1}{a}=5 \text {; }$
Thus identity $(4)$, becomes,
$a^{2}-1 / a^{ 2}=(+ - \sqrt{29})(5)$
$ \Rightarrow a^2-\frac{1}{a^2}=5( \pm \sqrt{29})$


 
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Question 164 Marks
If $a-\frac{1}{a}=8$ and $a \neq 0$ find:$(i)\ a+\frac{1}{a};(ii)\ a^2-\frac{1}{a^2}$
Answer
We know that,
$(a+b)^2=a^2+2 a b+b^2$
Given that $a-\frac{1}{a}=8$;
Substitute in equation $(1),$ we have
$ (8)^2=a^2+\frac{1^a}{a^2}-2$
$\Rightarrow a^2+\frac{1}{a^2}=64+2$
$\Rightarrow a^2+\frac{1}{a^2}=66$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=66+2$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=68 $
$\text { i) } a+\frac{1}{a}=\sqrt{68}$
$\Rightarrow \sqrt{17 \times 4}={ }_{-}^{+} 2 \sqrt{17}$
$\text { ii) } a^2-\frac{1}{a^2}=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)$
$\Rightarrow a^2-\frac{1}{a^2}={ }_{-}^{+} 2 \sqrt{17} \times 8$
$\Rightarrow a^2-\frac{1}{a^2}={ }_{-}^{+} 16 \sqrt{17}$
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Question 174 Marks
If $a+\frac{1}{a}=6$ and $a \neq 0$ find:$(i) a-\frac{1}{a};(ii) a^2-\frac{1}{a^2}$
Answer
We know that,
$(a+b)^2=a^2+2 a b+b^2$
and
$(a-b)^2=a^2-2 a b+b^2$
Thus,
$\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}+2 \dots.....(1)$
Given that $a+\frac{1}{a}=6$; Substitute in equation $(1),$ we have
$(6)^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow a^2+\frac{1}{a^2}=36-2$
$\Rightarrow a^2+\frac{1}{a^2}=34\ldots...(2)$
Similarly, consider
$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}-2$
$=34-2 [$ from $(2) ]$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=32$
$ \Rightarrow\left(a-\frac{1}{a}\right)= \pm \sqrt{32}$
$\Rightarrow\left(a-\frac{1}{a}\right)= \pm 4 \sqrt{2}\ldots....(3)$
$(ii)$ We need to find $a^2-\frac{1}{a^2}$
We know that, $a^2-\frac{1}{a^2}=\left(a-\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$a-\frac{1}{a}= \pm 4 \sqrt{2} ; a+\frac{1}{a}=6$
Thus,
$a^2-\frac{1}{a^2}=( \pm 4 \sqrt{2})(6)$
$ \Rightarrow a^2-\frac{1}{a^2}=( \pm 24 \sqrt{2})$
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[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip