If f(2x)= x^2 +4x+1 , what all values of t for which f((t/2)) = −((11)/4) where f represents a function


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Question Number 115854 by bemath last updated on 29/Sep/20

$I f f (2 x) = x^{2} + 4 x + 1, w h a t a l l$ $v a l u e s o f t f o r w h i c h f (\frac{t}{2}) = - \frac{11}{4}$ $w h e r e f r e p r e s e n t s a f u n c t i o n$
	Answered by bobhans last updated on 29/Sep/20
	$⇔ f(x)= ((x/2))^2 +4((x/2))+1 ⇔ f((t/2)) = ((t/4))^2 +4((t/4))+1 = −((11)/4) ⇒ (t^2 /(16)) + ((4t)/4) + ((15)/4) = 0 ⇒ t^2 +16t +60=0 ⇒ (t+6)(t+10)=0 → { ((t=−6)),((t=−10)) :}$
	$\Leftrightarrow f (x) = {(\frac{x}{2})}^{2} + 4 (\frac{x}{2}) + 1$ $\Leftrightarrow f (\frac{t}{2}) = {(\frac{t}{4})}^{2} + 4 (\frac{t}{4}) + 1 = - \frac{11}{4}$ $\Rightarrow \frac{t^{2}}{16} + \frac{4 t}{4} + \frac{15}{4} = 0$ $\Rightarrow t^{2} + 16 t + 60 = 0$ $\Rightarrow (t + 6) (t + 10) = 0 \to {\begin{cases} t = - 6 \\ t = - 10 \end{cases}$
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