Question

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

Answer 1

Heya!
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♦Quadratic Equations♦
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=> 2x² - x + ⅛ = 0


✔Simplify the Equation by taking LCM =>>


=> 16x² - 8x + 1 / 8 = 0


=> 16x² - 8x + 1 = 0


♦Now further it can be solved by two ways =>


$|1| .middle \: term \: splitting$

=> Splitting Factors ➡ -4 and -4


=> 16x² -4 x - 4x + 1 = 0


=> 4x ( 4x - 1 ) -1 ( 4x - 1 )

=> ( 4x - 1 ) ( 4x - 1 )

=> 4x = 1 , 4x = 1

=> x = ¼ , x = ¼



$|2| .by \: quadratic \: formula \: = > \\ \\ x = \frac{ - b \frac{ + }{ - } \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ \\ here \: = > b {}^{2} - 4ac \\ \\ = > ( - 8) {}^{2} - 4(16)(1) \\ \\ = 64 -6 4 \\ \\ = 0$


=> x = 8 / 32 , 8/32


=> 1/4 , 1/4



♦Equation has Real and Equal Roots !!


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Answer 2

❤❤here is your answer ✌ ✌ In attachment