Question

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

Answer 1

Hey there !!!!!!!!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

f(x) = x²+px+45 

If α and β are roots of f(x) then according to question 

 (α-β)² = 144

 α²+β²-2αβ = 144

Adding and subtracting 2αβ

 α²+β²+2αβ-4αβ = 144

 (α+β)² - 4αβ = 144

We know that if p(x) = ax²+bx+c has α,β as its roots then 

α+β = -b/a and αβ=c/a

So comparing x²+px+45 with ax²+bx+c

a=1 b= p c=45

(α+β)² - 4αβ = 144

= b²/a²-4c/a=144

=p²-4*45=144

p²=144+180

p²=324

p= +18 or -18

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hope this helped you..............