Question

If α and β are the zeroes of the quadratic polynomial f(x)= ax2+bx+c, then α-β=​

Answer 1

Answer:

$f \: (x) = {ax}^{2} + {bx}^{2} + c \:$

$\beta + \alpha \\ a \alpha + \: a \beta + b \:$

$\frac{ \beta (a \beta + b \: + a \alpha + b)}{(a \alpha + b) \: a \beta + b} \:$

$\frac{ a \beta + b \beta + a \alpha + b \alpha } { {a}^{2}a \beta + ab \alpha a \beta + {p}^{2} }$

observe, we have

α+β=

a

−b

αβ=

a

c

⇒ α

2

+β

2

=(α+β)

2

−2αβ

=

a

2

b

2

−

a

2c

Using equations on the right

=

a

2

(

a

c

)+ab(

a

−b

)+b

2

a(

a

2

b

2

−

a

2c

)+b(

a

−b

)

=

ac−b

2

+b

2

a

b

2

−2c−

a

b

2

=

ac

−2c

=

a

−2

Answer 2

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If α and β are zero of f(x)=ax2+bx+c. The evaluate α4×β4

Note that we have

= (α+β)4

=α4+β4+4C1α3β1+4C2α2β2+4C3α1β13

Then, α4+β4=(α+β)4−4(α3β1+α1β3)−6α2β2

= (α+β)4−4αβ(α2+β2)−6(αβ)2

Maxover, we have

α+β = −ab and αβ = ac

Then,

α4+β4=(−ab)4−4(ac)[(−ab)2−2(ac)]−6(ac)2

=a4b4−a4c(a2b2)+8(ac)2−6(ac)2

=a4b4−a34b2c+2a2C2