Find the discriminant of the equation 3x2 – 2x + 1 3 = 0 and hence find the nature of its roots. Find them, if they are real.
Step-by-Step Solution
Key Concept: For a quadratic equation $ax^2+bx+c=0$, the discriminant $D=b^2-4ac$ determines the nature of its roots: $D>0$ ⇒ two distinct real roots, $D=0$ ⇒ equal real roots, $D<0$ ⇒ two imaginary (complex conjugate) roots. The roots are given by $x=\frac{-b\pm\sqrt{D}}{2a}$.
1. Identify the coefficients\
The given quadratic is $3x^2-2x+13=0$, so\
\[ a=3,\quad b=-2,\quad c=13. \]
2. Compute the discriminant\
\[ D = b^2-4ac = (-2)^2-4\cdot3\cdot13 = 4-156 = -152. \]
3. Interpret the discriminant\
Since $D=-152<0$, the equation has no real roots; the roots are a pair of complex conjugates.
4. Find the roots (for completeness)\
Using the quadratic formula:\
\[ x = \frac{-b\pm\sqrt{D}}{2a} = \frac{-(-2)\pm\sqrt{-152}}{2\cdot3} = \frac{2\pm\sqrt{-152}}{6}. \]
Write $\sqrt{-152}=i\sqrt{152}=i\,2\sqrt{38}$, then\
\[ x = \frac{2\pm i\,2\sqrt{38}}{6}=\frac{1\pm i\sqrt{38}}{3}. \]
5. Conclusion\
The discriminant is $-152$, indicating two imaginary (complex) roots. The roots are $\displaystyle x = \frac{1}{3} \pm \frac{i\sqrt{38}}{3}$.
Hence, the equation has no real roots.
Correct Answer: Discriminant $D = -152$ (negative) ⇒ the roots are imaginary. The roots are $x = \frac{1}{3} \pm \frac{i\sqrt{38}}{3}$; therefore there are no real roots.