Question

find value of x if eqn is
$100 {x}^{2} - 20x + 1 = 0$
​

Answer 1

Answer:

$x = \frac{1}{10}$

Step-by-step explanation:

$100 {x}^{2} - 20x + 1 = 0 \\ 100 {x}^{2} - 10x - 10x + 1 = 0 \\ 10x(10x - 1) - 1(10x - 1) = 0 \\( 10x - 1)(10x - 1) = 0 \\ 10x - 1 = 0 \\ 10x = 1 \\ x = \frac{1}{10}$

Answer 2

Answer:

$\large x=\dfrac{1}{10}$

Step-by-step explanation:

Given :

$\large p(x)=100x^2-20x+1=0$

We have to find x

Rewrite given as

$\large p(x)=(10)^2-2\times10x\times1+(1)^2=0$

Now using identity here

$\large (a^2-2ab+b^2)=(a-b)^2$

$\large p(x)=(10x-1)^2=0\\\\\\\large (10x-1)=0\\\\\\\large (10x)=1\\\\\\\large x=\dfrac{1}{10}$

Thus we get answer.