Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(x)= ax^{2}+bx+c$,then evaluate
(v) $\alpha^{4}+ \beta^{4}$
(vi) $\frac{1}{a \alpha+b} + \frac{1}{a \beta+b}$
(vii) $\frac{\beta}{a \alpha+b} + \frac{\alpha}{a \beta+b}$
(viii) $a (\frac{\alpha^{2}}{\beta} + \frac{\beta^{2}}{\alpha}) + b (\frac{\alpha}{\beta} + \frac{\beta}{\alpha})$

Answer 1

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

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

α+β = −b/a …………………(1)

Product of the zeroes of the quadratic polynomial = constant term/ Coefficient of x²

αβ = c/a ……………………….(2)

SOLUTION OF (vi)  (vii)  and (viii)  ARE IN THE ATTACHMENT  :

(v) Given : α⁴ + β⁴

α⁴ + β⁴ = (α²)² + (β²)²

= (α² + β²)² –2α²β²

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

= ((α + β)² - 2αβ)² –(2αβ)² ………….. (3)

[(−b/a)² - 2(c/a)]² –[2(c/a)²]

From eq 1 & 2

= [b²/a² –2c/a]² - 2c²/a²

= [(b² - 2ac) /a²]²  -  2c²/a²

= (b² - 2ac)² /a⁴   -  2c²/a²

= [((b² - 2ac)²  -  2c²a²)  / a⁴]

Hence, the value of α⁴ + β⁴ is  [((b² - 2ac)²  -  2c²a²)  / a⁴] .

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Answer 2

Answer:

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(x)= ax^{2}+bx+c$,then evaluate

(v) $\alpha^{4}+ \beta^{4}$

(vi) $\frac{1}{a \alpha+b} + \frac{1}{a \beta+b}$

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

(viii) $a (\frac{\alpha^{2}}{\beta} + \frac{\beta^{2}}{\alpha}) + b (\frac{\alpha}{\beta} + \frac{\beta}{\alpha})$