Most important algebra questions asked in SSC CGL

In this article , some  important algebra  questions asked in SSC CGL previous year papers  are  discussed with explanations.

1. If x+\frac{1}{x}=\sqrt{2}  then find

(i) x^{16}+x^{12}+x^{8}+x^{4}+1

(ii) x^{60}+\frac{1}{x^{60}}

(iii) x^{61}+\frac{1}{x^{61}}

(iv) x^{99}+x^{95}+1

Sol.   Given x+\frac{1}{x}=\sqrt{2}  .

Squaring both sides  we get,  x^{2}+\frac{1}{x^{2}}+2=2

\Rightarrow x^{2}+\frac{1}{x^{2}}=0

\Rightarrow x^{4}+1=0. Hence x^{4}=-1

(i)  Now x^{16}+x^{12}+x^{8}+x^{4}+1= (x^{4})^{4}+ (x^{4})^{3}+(x^{4})^{2}+x^{4}+1

Putting   x^{4}=-1  , we get

x^{16}+x^{12}+x^{8}+x^{4}+1 =   (-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1


(ii) x^{60}+\frac{1}{x^{60}}=(x^{4})^{15}+\frac{1}{(x^{4})^{15}}=(-1)^{15}+\frac{1}{(-1)^{15}}=-1-1=-2

(iii) x^{61}+\frac{1}{x^{61}}= x^{60}\times x+\frac{1}{x^{60}\times x} = (x^{4})^{15}\times x+\frac{1}{(x^{4})^{15}\times x}

= (-1)^{15}\times x+ \frac{1}{(-1)^{15}\times x}=-x-\frac{1}{x}=-(x+\frac{1}{x})=-\sqrt{2}

(iv) x^{99}+x^{95}+1= x^{95}(x^{4}+1)+1=(x)^{95}(-1+1)+1= x^{95}\times 0+1=1

2.   If x+\frac{1}{x}=5 , then find the value of x^{3}+\frac{1}{x^{3}} .

Sol.  Cubing both the sides , we get  (x+\frac{1}{x})^{3}=125

\Rightarrow x^{3}+\frac{1}{x^{3}}+3\times x\times \frac{1}{x}(x+\frac{1}{x})=125           [\because  (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)  ]

\Rightarrow x^{3}+\frac{1}{x^{3}}+3\times 5=125                   [ putting the value of x+\frac{1}{x} ]

\Rightarrow  x^{3}+\frac{1}{x^{3}}=125-15.     Hence x^{3}+\frac{1}{x^{3}}=110


We can use the following formula directly  \bg_green \bg_green x+\frac{1}{x} =k , \, \, \, then\, \, \, \, \, x^{3}+\frac{1}{x^{3}}=k^{3}-3k

3. If x^{4}+\frac{1}{x^{4}}=194, x>0,    then  find   the value of  \left ( x-2 \right )^{2} .

Sol.  Given x^{4}+\frac{1}{x^{4}}=194

Adding 2 both sides we get x^{4}+\frac{1}{x^{4}}+2=194+2=196  . It  can be written as

\left ( x^{2} +\frac{1}{x^{2}}\right )^{2}=\left ( 14 \right )^{2}

Taking +ve square root of both sides, we get x^{2}+\frac{1}{x^{2}}=14

Again adding 2 both the sides we get  x^{2}+\frac{1}{x^{2}}+2=14+2=16

This implies    (x+\frac{1}{x})^{2}=4^{2}. Hence x+\frac{1}{x}=4 \, \, \, \, .......(i)

Now \left ( x-2 \right )^{2}=x^{^{2}}+4-4x =4x-1+4-4x=3   ( from equ. (i) x^{2}+1=4x)

Thus (x-2)^{2}=3

4. If  x^{99}+\frac{1}{x^{99}}=11   then find the value of x^{99}-\frac{1}{x^{99}}  .

Sol.  To solve the above question you  can use the following formula

(i ) If x^{n}+\frac{1}{x^{n}}=k then x^{n}-\frac{1}{x^{n}}=\sqrt{k^{2}-4}    and

(ii) If x^{n}-\frac{1}{x^{n}}=k  then x^{n}+\frac{1}{x^{n}}=\sqrt{k^{2}+4}

Proof. (i ) Given x^{n}+\frac{1}{x^{n}}=k

Squaring both sides we get x^{2n}+\frac{1}{x^{2n}}+2=k^{2}

\Rightarrow x^{2n}+\frac{1}{x^{2n}}=k^{2}-2.  Now subtracting 2 from both sides we get,  x^{2n}+\frac{1}{x^{2n}}-2=k^{2}-4

\Rightarrow (x^{n}-\frac{1}{x^{n}})^{2}=k^{2}-4

\Rightarrow x^{n}-\frac{1}{x^{n}}=\sqrt{k^{2}-4}

Similarly we can prove that  If x^{n}-\frac{1}{x^{n}}=k  then x^{n}+\frac{1}{x^{n}}=\sqrt{k^{2}+4}

Now using the above formula the value of  x^{99}-\frac{1}{x^{99}}   =\sqrt{11^{2}-4}=\sqrt{117}  .

5. If x^{2}-2\sqrt{5}\, x+1=0  , then find the value of x^{5}+\frac{1}{x^{5}}.

Sol.  Dividing the given equation by x, we get  x-2\sqrt{5}+\frac{1}{x}=0.  This implies  x+\frac{1}{x}=2\sqrt{5}  ………… (i)

On squaring both sides we get, (x+\frac{1}{x})^{2}=20. It gives x^{2}+\frac{1}{x^{2}}=20-2=18     ………………. (ii)

Now cubing both sides  of eqiuation (i), we get  x^{3}+\frac{1}{x^{3}}+3 \times x\times \frac{1}{x}(x+\frac{1}{x})=40\sqrt{5} . This implies x^{3}+\frac{1}{x^{3}}=40\sqrt{5}-6\sqrt{5}=36\sqrt{5}     …………….(iii)

squaring equation (ii) we get  x^{4}+\frac{1}{x^{4}}=(18)^{2}-2= 324-2=322    …………..(iv)

Now (x^{4}+\frac{1}{x^{4}})(x+\frac{1}{x})=322\times 2\sqrt{5} . This gives  x^{5}+x^{3}+\frac{1}{x^{3}}+\frac{1}{x^{5}}=644\sqrt{5}  . Thus


On putting the value of x^{3}+\frac{1}{x^{3}} from (iii),   the value of  x^{5}+\frac{1}{x^{5}}=644\sqrt{5}-34\sqrt{5}=610\sqrt{5}

6. If 30x^{2}-15x+1=0, then what is the value of 25x^{2}+\frac{1}{36x^{2}} .

Sol.   Given that  30x^{2}-15x+1=0 .

Dividing by 6x, we get 5x-\frac{15}{6}+\frac{1}{6x}=0 \Rightarrow 5x+\frac{1}{6x}=\frac{15}{6}. Now squaring both sides we get

25x^{2}+\frac{1}{36x^{2}}+2\times 5\times \frac{1}{6}=\frac{225}{36 }.     This implies  25x^{2}+\frac{1}{36x^{2}}=\frac{225}{36}-\frac{10}{6}=\frac{165}{36}=\frac{55}{12} .


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