QUADRATIC DOUBT :)

1. if a and b are integers and x2 + ax + b =0 has discriminant as a perfect square then prove that its roots are integers.

SOLUTION IS :
roots are { - a± √ ( a2 -4b)} / 2

if a2 -4b = 4m , then a will be even and roots will be integers

if a2 -4b= (2m +1)22 , then a will be odd and then also the roots will be integers.

tell me by this how can they predict if a will be odd or even ?

4 Answers

1
skygirl ·

i din get your question...........

1
skygirl ·

a^2 -4b = 4m

=> a^2 = 4b +4m

now if a is odd....

a^2 is odd. so LHS is odd.

but RHS is even.

1
playpower94 ·

sky my Q is how are they saying a is odd

NOT " IF a IS odd "

[2]

1
akash93 ·

(2m+1)2=odd
odd-even=odd
(2m+1)2-4b = odd
→a is odd

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