Question

The roots of the Quadratic equation
9x²-6x + 1 =0 are​

Answer 1

Answer:

Here is the answer.

Step-by-step explanation:

By factorisation method

${9x}^{2} - 6x + 1 = 0 \\ {9x}^{2} - 3x - 3x + 1 = 0 \\ 3x(3x - 1) - 1(3x - 1) = 0 \\ (3x - 1)(3x - 1) = 0 \\ 3x - 1 = 0 \: or \: 3x - 1 = 0 \\ 3x = 1 \: or \: 3x = 1 \\ x = \frac{1}{3} or \: x = \frac{1}{3}$

The roots of the quadratic equation is 1/3 and 1/3

Answer 2

$\mathfrak{{\pmb{{\underline{To~find}}:}}}$

• The roots of the quadratic equation $\sf{\pmb{9x^{2}-6x+1=0}}$

$\mathfrak{{\pmb{{\underline{Solution}}:}}}$

$\sf{~~~~~~~~~~ 9x^{2}-6x+1=0}$

$\boxed{\sf{S=-6~,~~P=9~~~~{\pmb{(-3,~-3)}}}}$

$\sf\implies{(9x^{2}-3x)+(-3x+1)=0}$

$\sf\implies{[3x(3x-1)~]+[-1(3x-1)~]=0}$

$\sf\implies{(3x-1)(3x-1)=0}$

$\sf\implies{3x-1=0~~~or~~~3x-1=0}$

$\sf\implies{x={\dfrac{1}{3}}~~~or~~~x={\dfrac{1}{3}}}$

$\therefore\mathfrak{{\pmb{{\underline{Required~answer}}:}}}$

Roots are :–

• $\underline{ \boxed{ \sf{ {\pmb{ \dfrac{1}{3}}} \: \: \: \: \: and \: \: \: \: \: {\pmb{ \dfrac{1}{3}}}}}}$