Question

please answer this question ​

Answer 1

Answer:

$\rm\huge\purple\star\underline\mathfrak\pink{SOLUTION :-.}$

Step-by-step explanation:

$= > {2x}^{2} + x + 4 = 0$

$= > 2( {x}^{2} + \frac{1}{2} x + 2) = 0$

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

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

$= > ( {x}^{2} + \frac{1}{2} x + \frac{1}{16} ) + 2 - \frac{1}{16} = 0$

$= > ( {x}^{2} + \frac{1}{2} + \frac{1}{6} ) + \frac{31}{16} = 0$

$= > {(x + \frac{1}{4}) }^{4} + \frac{31}{16} = 0$

$= > {(x + \frac{1}{4}) }^{4} = - \frac {31}{16}$

$\rm\mid\pink\star\underline\mathfrak\green{ Not \:possible\: as \:a \:square \:can't \:be \:negative \:therefore, It\: has \:no\: real \:roots.}$