Question

2x²-x+1/8=0 by grind roots by factorization
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Answer 1

$\huge \boxed{ \fcolorbox{cyan}{red}{Answer : }}$

●$\sf{2 {x}^{2} -x + 1/8 = 0}$

●$\sf{ {16x}^{2} -8x + 1= 0}$

●$\sf{ {16x}^{2} - 4x - 4x + 1 = 0}$

●$\sf{4x(4x -1) -1(4x-1) = 0}$

●$\sf{(4x-1) (4x-1)= 0}$

●$\sf{(4x-1)=0 \: or \: (4x-1)=0}$

$\bf{ \huge{ \boxed{ \red{ \tt{x = +1/4 and +1/4 \: }}}}}$

x = +1/4 and +1/4

Answer 2

Answer:

The roots are $\frac{1}{4}$ and $\frac{1}{4}$.

Step-by-step explanation:

The expression is:

$2x^{2}-x+\frac{1}{8}=0\\16x^{2}-8x+1=0$

Factorize the equation to compute the roots as follows:

$16x^{2}-8x+1=0\\16x^{2}-4x-4x+1=0\\4x(4x-1)-1(4x-1)=0\\(4x-1)(4x-1)=0$

Thus, the roots are $\frac{1}{4}$ and $\frac{1}{4}$.