110k views
1 vote
6r^2+6r+12=0
solve the square​

User Nayara
by
8.2k points

1 Answer

3 votes

Answer:


\large\boxed{\bold{NO\ REAL\ SOLUTION}}\\\text{in the set of complex numbers}\\\boxed{x=(-1\pm i\sqrt7)/(2(1))}

Explanation:


6r^2+6r+12=0\qquad\text{divide both sides by 6}\\\\(6r^2)/(6)+(6r)/(6)+(12)/(6)=0\\\\r^2+r+2=0\\\\\text{Use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{hen the equation has no real solution}\\\text{if}\ b^2-4ac=0,\ \text{hen the equation has one real solution}\ x=(-b)/(2a)\\\text{if}\ b^2-4ac>0,\ \text{hen the equation has two real solution}\ x=(-b\pm√(b^2-4ac))/(2a)


\text{We have}\ a=1,\ b=1,\ c=2.\\\\b^2-4ac=1^2-(4)(1)(2)=1-8=-7<0\\\\\bold{NO\ REAL\ SOLUTION}


\text{If you want solution in the set of complex numbers, then}\\\\√(-7)=√((-1)(7))=√(-1)\cdot\sqrt7=i\sqrt7\\\\x=(-1\pm i\sqrt7)/(2(1))\\\\x=(-1\pm i\sqrt7)/(2)

User Musako
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories