Question

If ${\alpha} {\beta}$ are the zeros of the polynomial $f(x)=ax^{2}+bx+c$,then $\frac{1}{\alpha^{2}}+ \frac{1}{\beta^{2}}$=
(a) $\frac{b^{2} -2ac} {a^{2}}$
(b) $\frac{b^{2} -2ac} {c^{2}}$
(c) $\frac{b^{2} +2ac} {a^{2}}$
(d) $\frac{b^{2} +2ac} {c^{2}}$

Answer 1

SOLUTION :

The correct option is (b) : (b² - 2ac)/c²

Given : α  and β are the zeroes of the  polynomial f(x) = ax² + bx + c .

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a ………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a  ……………………(2)

Given : 1/α²  + 1/β²

= (1/α  + 1/β)² - 2/αβ

[By using the identity : a² + b² =  (a + b)² - 2ab]

= ((α + β)/αβ)² - 2/αβ

= ((-b/a)/c/a)² - 2/( c/a)

[From eq 1 & 2]

=( - b/a × a/c)² - 2 × a/c

= (- b/c)² - 2a/c

= b² / c² - 2a/c

1/α²  + 1/β² = (b² - 2ac)/c²

Hence, the value of 1/α²  + 1/β² is  (b² - 2ac)/c² .

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

(b)is right answer ....