Question

Factorise : 21x^2 - 2x + 1/21

Answer 1

$21 x^{2}-2 x+\frac{1}{21}=\frac{1}{21}, \frac{1}{21}$

Step-by-step explanation:

To find the roots of the quadratic equation equate it to 0

Let $f(x)=21 x^{2}-2 x+\frac{1}{21}=0$

On taking LCM, we get,

$441 x^{2}-42 x+1=0$

The above in equation is in $(a-b)^{2} = 0$ form

Where, a = 21x; b = 1

On using, the above formula, we get,

$(21 x-1)^{2}=0$

On solving, we get,

$21 x-1=0; \ \ 21 x-1=0$

$\Rightarrow 21 x=1; \ \ 21 x=1$

Thus, the value of x are,

$x=\frac{1}{21}, \frac{1}{21}$

Answer 2

Answer:

$21x^{2}-2x+\frac{1}{21}}\\=\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)$

Step-by-step explanation:

$Given\\ quadratic \: expression:\\21x^{2}-2x+\frac{1}{21}$

$=\left(\sqrt{21}x\right)^{2}-2x+\left(\frac{1}{\sqrt{21}}\right)^{2}$

$=\left(\sqrt{21}x\right)^{2}-2\times \sqrt{21} x \times \frac{1}{\sqrt{21}}+\left(\frac{1}{\sqrt{21}}\right)^{2}$

$=\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)^{2}$

/* By algebraic identity:

a²-2ab+b² = (a-b)² */

$=\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)$

Therefore,

$21x^{2}-2x+\frac{1}{\sqrt{21}}\\=\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)\big(\sqrt{21} x-\frac{1}{\sqrt{21}}\big)$

•••♪