Question

If the square of difference of the zeroes of the quadratic polynomial f(x)= x2+px+45 is equal to 144,
find the value of p.

Answer 1

Given quadratic polynomial,
f(x)=x²+px+45
Let A and B be the zeroes of polynomial f(x).
sum of the zeroes=A+B=-p
A+B=-p.............................................1
and,
product of zeroes=AB=45..........2
As per given condition we have,
(A-B)²=144

A²+B²-2AB=144

(A+B)²-4AB=144
putting eq1 and eq2 in above equation we get,
(-p)²-4(45)=144

p²-180=144

p²=324

p=±√324

p=±18


Hence value of p=±18.

Answer 2

$\huge\mathtt{\fbox{\red{Answer}}}$

Given quadratic polynomial,

f(x)=x²+px+45

Let A and B be the zeroes of polynomial f(x).

sum of the zeroes=A+B=-p

A+B=-p.............................................1

and,

product of zeroes=AB=45..........2

As per given condition we have,

(A-B)²=144

A²+B²-2AB=144

(A+B)²-4AB=144

putting eq1 and eq2 in above equation we get,

(-p)²-4(45)=144

p²-180=144

p²=324

p=±√324

p=±18

Hence value of p=±18.