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Algebra Level 1

Find all the real roots to the equation x 2 + x + 1 = 0. x^2+x+1=0.

No real roots x = − 1 x=-1 x = − 0.5 x=-0.5 x = 1 x=1

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2 solutions

Rishabh Jain
Jan 30, 2016

x 2 + x + 1 = ( x + 1 2 ) 2 + 3 4 \large x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4} Since square of any real quantity is greater than zero ⇒ ( x + 1 2 ) 2 + 3 4 ≥ 3 4 \large\Rightarrow (x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4} and hence given expression is non zero for real values of x.

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I did it by the quadratic formula but this a really unique and intuitive way!!!! Nice job👍👍👍

Racchit Jain - 5 years, 4 months ago

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This is the basic reasoning (Completing the squares is doing the job for us) : D

Rishabh Jain - 5 years, 4 months ago
Akshat Sharda
Jan 30, 2016

x 2 + x + 1 = 0 b 2 − 4 a c = 1 2 − 4 ( 1 ) ( 1 ) = − 3 < 0 ⇒ Roots are imaginary. x^2+x+1=0 \\ b^2-4ac=1^2-4(1)(1)=-3<0 \\ \Rightarrow \text{Roots are imaginary.}

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I solved with this way

A.M .O - 1 year, 1 month ago

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