Given-the-roots-of-the-quadratic-equation-4x-2-4x-5-0-are-and-f-x-is-a-quadratic-function-where-f-f-and-f-0-6-Find-f-x-

Question Number 163323 by ZiYangLee last updated on 06/Jan/22

Given the roots of the quadratic equation

4 x^{2} - 4 x + 5 = 0 are α and β .

f (x) is a quadratic function where f (α) = β,

f (β) = α and f (0) = 6 .

Find f (x) .

Commented by cortano1 last updated on 06/Jan/22

l e t f (x) = {p x}^{2} + q x + 6

\Rightarrow {\begin{cases} f (α) = p α^{2} + q α + 6 = β \\ f (β) = p β^{2} + q β + 6 = α \end{cases}

(1) + (2) \Rightarrow p [{(α + β)}^{2} - 2 α β] + q (α + β) + 12 = α + β

\Rightarrow p [1 - \frac{5}{2}] + q (1) + 12 = 1

\Rightarrow - \frac{3}{2} p + q = - 11

\Rightarrow - 3 p + 2 q = - 22

\Rightarrow q = \frac{3 p - 22}{2}

∴ f (x) = {p x}^{2} + (\frac{3 p - 22}{2}) x + 6

Answered by Rasheed.Sindhi last updated on 06/Jan/22

f (x) = {a x}^{2} + b x + c (s a y)

f (0) = c = 6

f (α) = a α^{2} + b α + 6 = β \dots \dots \dots \dots . (i)

f (β) = a β^{2} + b β + 6 = α \dots \dots \dots \dots . . (i i)

(i) + (i i) :

a (α^{2} + β^{2}) + b (α + β) + 12 = α + β

a {{(α + β)}^{2} - 2 α β} + b (α + β) + 12 = α + β

4 x^{2} - 4 x + 5 = 0 \Rightarrow {\begin{cases} α + β = 1 \\ α β = \frac{5}{4} \end{cases}

a {{(1)}^{2} - 2 (\frac{5}{4})} + b (1) + 12 = 1

a (- \frac{3}{2}) + b = - 11

3 a - 2 b = 22 \dots \dots \dots \dots \dots \dots \dots A

(i) - (i i) :

f (α) - f (β) = a (α^{2} - β^{2}) + b (α - β) = - (α - β)

= a (α + β) + b = - (α + β); α \neq β

a (1) + b = - (1)

a + b = - 1 \dots \dots \dots \dots \dots \dots \dots . B

A & B :

3 a - 2 (- 1 - a) = 22

3 a + 2 + 2 a = 22

a = 4 \Rightarrow b = - 5

f (x) = 4 x^{2} - 5 x + 6

Commented by peter frank last updated on 06/Jan/22

great

Commented by Rasheed.Sindhi last updated on 06/Jan/22

T h a n k y o u P e t e r s i r!

Commented by Tawa11 last updated on 06/Jan/22

Great sir

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