Question

If the roots of the equation x^2+bx+c=0 are two consecutive integers, what is the value of b^2-4c-1

Answer 1

hi
do you know
sum of roots for \(ax^2+bx+c=0\)
is -b/a

and product of roots =c/a
?

Answer 2

no..i didnt! Thanks for that. let me try now and see if I will get stuck

Answer 3

you will also need this :
\((p-q)^2 = (p+q)^2 -4pq\)

p and q being the roots of the equation

Answer 4

and since roots are consecutive
p-q =\(\pm 1\)

Answer 5

umm...can you just take me through step by step

Answer 6

in this case
a=1
right ?

Answer 7

so, p+q = -b
pq = c
makes sense till now ?

Answer 8

lot of sense !

Answer 9

what does p+q represent

Answer 10

p and q are the roots of the equation

so p+q = sum of the roots = -b/a =-b

Answer 11

ok...i get it now

Answer 12

can you try to continue ?
use the formula i gave above

p-q =1
pq =c
p+q =-b

Answer 13

let me show you how I have understood and interpreted.
if the roots are (m) and (m+1), where m is an integer then:
sum of roots= (m)+(m+1)=2m+1=-b
product of roots= (m)(m+1)= \[m ^{2}+m=c \]
so \[b ^{2}=(-b)^{2}=[-(2m+1)]^{2}= 4m ^{2}+4m+1\]
Therefore
\[b ^{2}-4c-1=(4m ^{2}+4m+1)-4(m ^{2}+m)-1\]

am i on the right track?

Answer 14

absolutely!
you just did it the other way , but you will get the correct answer after you simplify :)

Answer 15

yaaaaay! thanks

Answer 16

welcome ^_^