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Please help :) Describe the nature of the roots for this equation. 3x^2+x-5=0

ATwo complex roots
BTwo real, rational roots
COne real, double root
DTwo real, irrational roots

Verified Answer :

naǫ

\(3x^2+x-5=0 \\ \\ a=3 \\ b=1 \\ c=-5 \\ \Delta=b^2-4ac=1^2-4 \times 3 \times (-5)=1+60=61\)

The discriminant Δ is greater than 0, so there are two real roots.

\(x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-1 \pm \sqrt{61}}{2 \times 3}=\frac{-1 \pm \sqrt{61}}{6}\)

√61 is an irrational number, so both roots will be irrational.

The answer is D.

ZorosPaPa

Answer: D

Step-by-step explanation: A-pex

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