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quadratic polynomial P(x)=ax^2+bx+c has positve coefficients a,b,c in ap.P(x)=0 has integer roots p,q.Find p+q+pq

```
2 years ago

```							Dear Meenakshi

Since a, b & c are in Arithmetic Progression, b-a = c-b = Constant = K
Also p & q are the two integer roots of of the equation ax^2+bx+c = 0 — (1)
then p + q = -b/a and pq = c/a,
Now the equation can be written as x^2-(p+ q) +p q = 0 — (2)
Comparing Eq. (1) and (2) a = 1, b = -(p+ q) and c = p q
K = c - b = p q - (-p- q) = pq + p+q — (3)
K = b - a = -p - q -1 = -(p+q+1) — (4)
Therefore a = 1, b = K + 1, c = 2K +1
Now substitute the possible values of K, given in the Answers 3,5,7, 14
if K=3, x^2+4x+7, -(p+q) = 4 & pq = 7, is prime number and no integer roots.
if K=5, x^2+6x+11, -(p+q) = 6 & pq =11, is prime number and no integer roots.
If K=7, x^2+8x+15, -(p+q) = 7 & pq =15, then the roots are p= -3 & q= -5.
if K=14, x^2+15x+29, -(p+q) = 15 & pq =29, is prime number and no integer roots.
Therefore K = -p - q -1 = 3 + 5 -1 = 7,
Regards

```
2 years ago
```							p+q =-b/a, pq=c/a p and q are interger roots so there addition and multiplication will also be integers.So a is factor of b and cLet b =a×m and c= a×n, n,m€RP+q= -m , pq= na,b,c are in APSo, b-a=c-b am-a=an-amm-1=n-mNow p+q+pq= -m+n or n-m=m-1We have options 3,5,7,141) if m-1 =3 then m=4 and n=7, 7 is prime no. And multiplicaton of two no.s can`t be a prime no. So, this option is incorrect2) if m-1=5 then m=6 and n=11, prime no. So option is incorrect3) if m-1=7 then m=8 and n=15, not a prime no. So, this can be correct option4) if m-1=14 then m=15 and n=29, prime no. , Incorrect optionOnly option C is satisfying
```
2 years ago
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