A number theory problem by Aditya Raj

Let two zeros are α and β where α > β According given condition (α - β)2 = 144 Let p(x) = x2 + px + 45 α + β = − b = − p = - p a 1

αβ = c = 45 = 45 a 1

now (α - β)2 = 144 (α + β)2 – 4 αβ = 144 (-p)2 – 4 (45) = 144 Solving this we get p = ± 18

The roots are $x = \frac{-p \pm\sqrt{p^2 - 180}}{2}$ , and the squared difference of these roots computes to:

$[\frac{-p + \sqrt{p^2 - 180}}{2} - \frac{-p - \sqrt{p^2 - 180}}{2}]^2 = 144;$

or $[\sqrt{p^2-180}]^2 = 144;$

or $p^2-180=144;$

or $p^2 = 324;$

or $\boxed{p=18}.$

Hello,

as stated that the squared difference of zeros is 144,

for the quadratic x^(2) + px + 45, by looking at c = 45,

it has factors of [1,3,5,9,15],

as by combinatination, it has

(x+1)(x+45) ----> x=-1, x=-45 squared difference = [ -1-(-45)]^2 = [-45-(-1)]^2= 1936

(x+3)(x+15) ---> x=-3 , x=-15 squared difference = [ -3-(-15)]^2 = [ -15-(-3)]^2 = 144

(x+5)(x+9) ---> x=-5 , x=-9 squared difference = [-5-(-9)[^2 = [ -9-(-5)]^2 =16

stated that the squared difference is 144,so automatically it is (x+3)(x+15),

by comparison x^(2) + 18x + 45 = x^(2) + px +45

px = 18x

p = 18,

therefore p = 18....

thanks...