Given the roots of the quadratic equation 4x^2 −4x+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).


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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
$let f(x)= px^2 +qx+6 ⇒ { ((f(α)=pα^2 +qα+6=β)),((f(β)=pβ^2 +qβ+6=α)) :} (1)+(2)⇒p[(α+β)^2 −2αβ ]+q(α+β)+12=α+β ⇒p[ 1−(5/2) ]+q(1)+12=1 ⇒−(3/2)p+q=−11 ⇒−3p+2q=−22 ⇒q=((3p−22)/2) ∴ f(x)= px^2 +(((3p−22)/2))x+6$
$l e t f (x) = {p x}^{2} + q x + 6$ $\Rightarrow {\begin{cases} f (α) = p α^{2} + q α + 6 = β \\ f (β) = p β^{2} + q β + 6 = α \end{cases}$ $(1) + (2) \Rightarrow p [{(α + β)}^{2} - 2 α β] + q (α + β) + 12 = α + β$ $\Rightarrow p [1 - \frac{5}{2}] + q (1) + 12 = 1$ $\Rightarrow - \frac{3}{2} p + q = - 11$ $\Rightarrow - 3 p + 2 q = - 22$ $\Rightarrow q = \frac{3 p - 22}{2}$ $∴ f (x) = {p x}^{2} + (\frac{3 p - 22}{2}) x + 6$
	Answered by Rasheed.Sindhi last updated on 06/Jan/22
	$f(x)=ax^2 +bx+c (say) f(0)=c=6 f(α)=aα^2 +bα+6=β.............(i) f(β)=aβ^2 +bβ+6=α..............(ii) (i)+(ii): a(α^2 +β^2 )+b(α+β)+12=α+β a{(α+β)^2 −2αβ}+b(α+β)+12=α+β 4x^2 −4x+5=0⇒ { ((α+β=1)),((αβ=(5/4))) :} a{(1)^2 −2((5/4))}+b(1)+12=1 a(−(3/2))+b=−11 3a−2b=22.....................A (i)−(ii): f(α)−f(β)=a(α^2 −β^2 )+b(α−β)=−(α−β) =a(α+β)+b=−(α+β) ;α≠β a(1)+b=−(1) a+b=−1......................B A & B: 3a−2(−1−a)=22 3a+2+2a=22 a=4⇒b=−5 f(x)=4x^2 −5x+6$
	$f (x) = {a x}^{2} + b x + c (s a y)$ $f (0) = c = 6$ $f (α) = a α^{2} + b α + 6 = β . . . . . . . . . . . . . (i)$ $f (β) = a β^{2} + b β + 6 = α . . . . . . . . . . . . . . (i i)$ $(i) + (i i) :$ $a (α^{2} + β^{2}) + b (α + β) + 12 = α + β$ $a {{(α + β)}^{2} - 2 α β} + b (α + β) + 12 = α + β$ $4 x^{2} - 4 x + 5 = 0 \Rightarrow {\begin{cases} α + β = 1 \\ α β = \frac{5}{4} \end{cases}$ $a {{(1)}^{2} - 2 (\frac{5}{4})} + b (1) + 12 = 1$ $a (- \frac{3}{2}) + b = - 11$ $3 a - 2 b = 22 . . . . . . . . . . . . . . . . . . . . . A$ $(i) - (i i) :$ $f (α) - f (β) = a (α^{2} - β^{2}) + b (α - β) = - (α - β)$ $= a (α + β) + b = - (α + β); α \neq β$ $a (1) + b = - (1)$ $a + b = - 1 . . . . . . . . . . . . . . . . . . . . . . B$ $A & B :$ $3 a - 2 (- 1 - a) = 22$ $3 a + 2 + 2 a = 22$ $a = 4 \Rightarrow b = - 5$ $f (x) = 4 x^{2} - 5 x + 6$
	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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