September 10, 2024

# 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,  $x^{2}+\frac{1}{x^{2}}+2=2$

$\Rightarrow&space;x^{2}+\frac{1}{x^{2}}=0$

$\Rightarrow&space;x^{4}+1=0$. Hence $x^{4}=-1$

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

Putting   $x^{4}=-1$  , we get

$x^{16}+x^{12}+x^{8}+x^{4}+1$

=   $(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$

=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) $x^{61}+\frac{1}{x^{61}}=&space;x^{60}\times&space;x+\frac{1}{x^{60}\times&space;x}$

= $(x^{4})^{15}\times&space;x+\frac{1}{(x^{4})^{15}\times&space;x}$

= $(-1)^{15}\times&space;x+&space;\frac{1}{(-1)^{15}\times&space;x}$

=$-x-\frac{1}{x}=-(x+\frac{1}{x})=-\sqrt{2}$

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

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  $(x+\frac{1}{x})^{3}=125$

$\Rightarrow&space;x^{3}+\frac{1}{x^{3}}+3\times&space;x\times&space;\frac{1}{x}(x+\frac{1}{x})=125$

[$\because$  $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
]

$\Rightarrow&space;x^{3}+\frac{1}{x^{3}}+3\times&space;5=125$                   [ putting the value of $x+\frac{1}{x}$

]

$\Rightarrow$  $x^{3}+\frac{1}{x^{3}}=125-15$

.     Hence $x^{3}+\frac{1}{x^{3}}=110$

OR

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

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

.

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

$\left&space;(&space;x^{2}&space;+\frac{1}{x^{2}}\right&space;)^{2}=\left&space;(&space;14&space;\right&space;)^{2}$

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 $x+\frac{1}{x}=4&space;\,&space;\,&space;\,&space;\,&space;.......(i)$

Now $\left&space;(&space;x-2&space;\right&space;)^{2}=x^{^{2}}+4-4x$

$=4x-1+4-4x$=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 $x^{2n}+\frac{1}{x^{2n}}+2=k^{2}$

$\Rightarrow&space;x^{2n}+\frac{1}{x^{2n}}=k^{2}-2$

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

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

$\Rightarrow&space;x^{n}-\frac{1}{x^{n}}=\sqrt{k^{2}-4}$

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 $x^{2}-2\sqrt{5}\,&space;x+1=0$

, 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  $x^{3}+\frac{1}{x^{3}}+3&space;\times&space;x\times&space;\frac{1}{x}(x+\frac{1}{x})=40\sqrt{5}$ . This implies $x^{3}+\frac{1}{x^{3}}=40\sqrt{5}-6\sqrt{5}=36\sqrt{5}$

…………….(iii)

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

Now $(x^{4}+\frac{1}{x^{4}})(x+\frac{1}{x})=322\times&space;2\sqrt{5}$

. This gives  $x^{5}+x^{3}+\frac{1}{x^{3}}+\frac{1}{x^{5}}=644\sqrt{5}$  . Thus

$x^{5}+\frac{1}{x^{5}}=644\sqrt{5}-(x^{3}+\frac{1}{x^{3}})$

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 $5x-\frac{15}{6}+\frac{1}{6x}=0$ $\Rightarrow&space;5x+\frac{1}{6x}=\frac{15}{6}$
. Now squaring both sides we get

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

.

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#### Bina singh

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