SOLUTION: 9x^2+30x+25=11

Algebra ->  Algebra  -> Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 9x^2+30x+25=11       Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Complex Numbers Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Complex Numbers Question 28750: 9x^2+30x+25=11 Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!9x^2+30x+25=11 9x^2+30x+25-11=0 9x^2+30x+14=0 ----(1) Given ax^2 +bx +c = 0, by formula x = [(-b)+ orminus sqrt(b^2-4ac)]/2a Here a=9, b=30 and c-14 x= [(-30)+or minus sqrt(30^2-4X9X14)]/(2X9) = [(-30)+or minus sqrt(900-504)]/(18) = [(-30)+or minus sqrt(396)]/(18) = [(-30)+or minus sqrt(36X11)]/(18) = [(-30)+or minus 6sqrt(11)]/(18) = [-5+orminus rt(11)]/3 (cancelling 6 in the nr and in the dr) Therefore x= (-5+sqrt(11)]/3 or x= (-5-sqrt(11)]/3 Verification: Putting x=(-5+sqrt(11)]/3 in (1) LHS = 9x^2+30x+14 = [-5+rt(11)]^2+10[-5+rt(11)]+14 =(-5)^2+ (rt(11))^2-10rt(11)-50+10rt(11)+14 =25+11-50+14 =50-50=0 =RHS Note: Since quadratic surds occur in conjuagate pairs the other value also must hold