Question

x²+1/4x+1/4=0
plzz solve this equation by steps..​

Answer 1

Answer:

Step-by-step explanation:

$x^2 + \frac{1}{4} x + \frac{1}{4} = 0\\\\(a = 1, b= \frac{1}{4}, c = \frac{1}{4})\\Quadratic formula: x = \frac{-b +- \sqrt{b^2-4ac}}{2a} \\x = \frac{-\frac{1}{4} +- \sqrt{\frac{1}{4}^2-4*1*\frac{1}{4}}}{2*1} \\\\x = \frac{-\frac{1}{4} +- \sqrt{\frac{1}{16}-1}}{2}$

$x =\frac {\frac{-1}{4 }+-\sqrt{\frac{-15}{16} } }{2}$

Since discriminant < 0, therefore no real roots exist for the equation

For imaginary roots,

$x =\frac {\frac{-1}{4 }+-\sqrt{\frac{-15}{16} } }{2}\\\\\\x =\frac {\frac{-1}{4 }+-\sqrt{\frac{15}{16}*(-1) } }{2}\\\\\\\\x =\frac {\frac{-1}{4 }+-\sqrt{\frac{15}{16}*(i^2) } }{2}\\\\\\\\x =\frac {\frac{-1}{4 }+-(i) \sqrt{\frac{15}{16} } }{2}\\\\\\\\x = \frac{\frac{-1}{4}+- i \frac{\sqrt{15} }{4} }{2}\\x = \frac{-1 +- i \sqrt{15} }{2*4}$

Now the roots are:

$x = \frac{-1+i\sqrt{15} }{8} and x = \frac{-1-i\sqrt{15} }{8}$

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