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
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