IIT JEE , AIEEE Forum - Mathematics , Algebra

sinjan.j

if one of the roots of the equation x^2 + ax + 8 is 4 while the equation

x^2 + ax + b has equal roots. find b.

sinjan.j

i can't understand the meaning of the question....

please help as fast as possible..!!

tango_goat

one root is 4 so other root is 2.....produc of roots=8
a=-6
so eqn with equal roots has sum of roots=-6
so eqn is x^2-6*x+b
b=9...equal roots are 3

nadeemoidu

one of the roots of the equation x^2 + ax + 8 is 4
So 4^2+ 4a +8=0
=> 24 + 4a =0
=> a=-6

Again substitute in the 2nd equation.
x^2 - 6x +b=0

The roots will be equal if D=0
i.e 6^2 - 4b = 0
b=9

nivedh_89

4 is one of the roots of x^2 + ax + 8 = 0...............
therefore,
4^2+4a+8 = 0...........................
a = -6............!!!!!!!!!!

now,
x^2 + ax + b = 0..................that is...........
x^2 - 6x + b = 0..............
this equation has equal roots(say 'm')....................!!!!!!

therefore,
2m = 6.......m = 3....................!!!!!!!!!
m^2 = b................
therefore,
b = 9...........!!!!!!!!

splashprabhu

its a simple question.

prod. of roots = 4*x₁ = 8
x₁ =2
sum of roots= -a = 4 + 2 = 6
a = -6
equal roots means its a perfect square
b = (a/2)²

sinjan.j

splashprabhu's answer is ver nice...!!!
i liked it..!!1

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