To factorise the quadratic polynomial
x 2+bx+c=0 where b and c are positive and represent it geometrically
Answers
Answer:
If a,b,c are in arithmetic progression, then 2b=a+c
α+β+αβ= (sum of roots) + (product of roots)
⇒α+β+αβ=−
a
b
+
a
c
=
a
c−b
=
a
b−a
=
a
b
−1
α+β+αβ is an integer ⇒a∣b (a divides b)
2b=a+c⇒a∣c
Write (a,b,c) as (a,a(1+r),a(1+2r))
Now, determinant of p(x) should be a square since roots are integers.
D=b
2
−4ac=a
2
(1+r)
2
−4a⋅a(1+2r)=a
2
[(1+r)
2
−4(1+2r)]
⇒(1+r)
2
−4(1+2r)=k
2
⇒r
2
−6r−(3+k
2
)=0
This is a new quadratic equation in r. r is also an integer, hence its determinant is also a square.
D
′
=36−4[−(3+k
2
)]=4(12+k
2
)
⇒12+k
2
=p
2
⇒(p−k)(p+k)=12
Looking for integral solutions,
(k,p)∈{(2,4),(−2,−4),(2,−4),(−2,4)}
⇒k=±2
⇒k
2
=4
r=7 or r=−1
For r=−1, D is negative. Not possible.
For r=7 , the equation is ax
2
+8ax+15a=0. Its roots are −3 and −5.
Hence, α+β+αβ=7