Question

Find the roots of the following quadratic equation by factorisation

$100x2 - 20x + 1 = 0$​

Answer 1

Answer:

x= 1/10 or x = 1/10

Step-by-step explanation:

100x^2 -20x + 1

so here the sum should be 20 and product should be 100

the possible pair is -

-10×-10=100

-10-10 = -20

Hence,

100x^2 - 10x - 10x + 1 = 0

by taking common

10x ( 10x - 1 ) - 1( 10x - 1 ) = 0

(10x - 1) ( 10x - 1 ) = 0

x= 1/10 or x = 1/10

hope u understood

pls mark brainlist

Answer 2

$\Large\textrm{Solution}$

We will take substitution twice here.

$\;$

$\textrm{Let us consider a substitution.}$

$\rm{t=10x}$

$\;$

$\textrm{By squaring both sides,}$

$t^{2}=100x^{2}$

$\;$

$\textrm{Eventually we get,}$

$\cdots\longrightarrow\boxed{\rm{t^{2}-2t+1=0}}$

$\;$

$\textrm{Now,}$

$\;$$\rm{(t-1)^{2}=0}$

$\;$

$\textrm{By taking resubstitution,}$

$\rm{(10x-1)^{2}=0}$

$\rm{\therefore x=\dfrac{1}{10}\ (Double\ solution)}$

$\;$

$\Large\textrm{Learn More}$

Consider $\rm{f(x)}$ has a total of $\rm{n}$ roots $y_{1},y_{2},\cdots,y_{n}$, and let,

$\rm{f(x)=a_{0}x_{n}+a_{1}x_{n-1}+\cdots+a_{n-1}x_{1}+a_{n}x_{0}}$

and

$\rm{f(x)=(x-y_{1})(x-y_{2})\cdots(x-y_{n})}$

$\;$

Comparing this to the following equation,

$\rm{x^{n}\cdot f(\dfrac{1}{x})=a_{0}x_{n}+a_{1}x_{n-1}+\cdots+a_{n-1}x_{1}+a_{n}x_{0}}$

and

$\rm{x^{n}\cdot f\left(\dfrac{1}{x}\right)=(1-y_{1}x)(1-y_{2}x)\cdots(1-y_{n}x)}$

Now, this result shows that the equation having reciprocal roots is obtained by rearranging the order of the coefficients.

$\;$

For example where $ca\neq 0$,

$cx^{2}+bx+a=0$ has reciprocal roots of $ax^{2}+bx+c=0$.

This information may be helpful, when we deal with reciprocal values such as $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$, etc.