Question

$2x { = }^{2} x + \frac{1}8{} = 0$
​

Answer 1

Step-by-step explanation:

Given : 2x^2-x+\frac{1}{8}=02x

2

−x+

8

1

=0

To Find: x

Solution:

2x^2-x+\frac{1}{8}=02x

2

−x+

8

1

=0

16x^2-8x+1=016x

2

−8x+1=0

16x^2-4x-4x+1=016x

2

−4x−4x+1=0

4x(4x-1)-1(4x-1)=04x(4x−1)−1(4x−1)=0

(4x-1)(4x-1)=0(4x−1)(4x−1)=0

Answer 2

Answer

$given \: = {2x}^{2} - x + \frac{1}{8} = 0 \\ to \: find = </u><u>x</u><u> \\ </u><u>solution</u><u> </u><u>:</u><u> \: \\ {2x}^{2} - x + \frac{1}{8} = 0 \\ \\ {16x}^{2} - 8x + 1 = 0 \\ {16x}^{2} - 4x - 4x + 1 = 0 \\ \\ 4x(4x - 1) - 1(4x - 1) = 0 \\ \\ (4x - 1)(4x - 1) = 0 \\ \\ x = \frac{1}{4} \: </u><u>,</u><u> \frac{1}{4}$

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