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

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

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 $5x-\frac{15}{6}+\frac{1}{6x}=0$ $\Rightarrow 5x+\frac{1}{6x}=\frac{15}{6}$

. Now squaring both sides we get

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