Question

If the squared difference of the zeroes of the quadratic polynomial f (x)=x^2+px+45 is equal to 144 then find the value of p

Answer 1

We have,

f(x)=x²+px+45

Let @ and ß be the zeros of the above polynomial

Here,

@+ß= -p and @ß=45

Given:

(@-ß)²=144................[1]

Firstly,

@²+ß²

=(@+ß)²-2@ß

=p²-2(45)

=p²-90

Expanding [1],

(@-ß)²=144

→@²+ß²-2@ß=144

→(p²-90)-2(90)=144

→p²-180=144

→p²=324

→p= ±18

For p= ±18,the given polynomial satisfies the given condition

Answer 2

ANSWER:

+18 or -18

METHOD:

Given polynomial is

$\bf \: f(x) = {x}^{2} + px + 45$

It is asked to find the value of p

There is another information also given in the question that

squared difference of the zeroes of polynomial is 144

Let us assume that

m,n are the roots of given f(x)

let us find

Sum of roots and product of roots

SUM OF ROOTS:

$m + n= \dfrac{ - b}{a}$

$= > m + n = \dfrac{ - p}{ 1} = - p........(1)$

PRODUCT OF ROOTS:

$mn = \dfrac{c}{a}$

$mn = \dfrac{45}{1} = 45......(2)$

Now

as per the given statement

we can write as

${(m - n)}^{2} = 144.....(3)$

let us first find

${m}^{2} + {n}^{2} = {(m + n)}^{2} - 2mn$

$= { ( - p)}^{2} - 90 = {p}^{2} - 90$

substituting all the values in equation 3

${(m - n)}^{2} = 144$

${m}^{2} + {n}^{2} - 2mn = 144$

${p}^{2} - 90 - 90 = 144 \\ {p}^{2} = \sqrt{324}$

$\bf \boxed{ p = +1 8 \: or \: - 18}$

FORMULAE USED:

$= > sum \: of \: roots = \dfrac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$=>product \: of \: roots = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$