Math, asked by omanshsahu03, 1 month ago

To factorise the quadratic polynomial
x 2+bx+c=0 where b and c are positive and represent it geometrically​

Answers

Answered by logavignesh18
1

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

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