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Would anyone be able to show me how to solve the question below?

Given the function $$f(x)=4\sqrt x+\sqrt 2$$

Find $t$ such that $$f(t)=3\sqrt 2$$

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  • $\begingroup$ This isn't clear, at least not to me. Are you integrating something? What? You say you want a root...which suggests that you are trying to solve something like $f(x)=0$, but I can't guess what function you have in mind. $\endgroup$ – lulu Oct 8 '16 at 14:47
  • $\begingroup$ Ah, I see from the edit that you are just trying to find $x$ such that $4\sqrt x+\sqrt 2 = 3\sqrt 2$. Poor notation...I will edit your post to clarify. $\endgroup$ – lulu Oct 8 '16 at 14:50
  • $\begingroup$ Now, where do you get stuck? $4\sqrt t+\sqrt 2 = 3\sqrt 2 \implies 4\sqrt t=2\sqrt 2$. Does that help? $\endgroup$ – lulu Oct 8 '16 at 14:52
  • $\begingroup$ I got there although I'm not sure what to do after that? $\endgroup$ – Jude Molloy Oct 8 '16 at 14:55
  • $\begingroup$ Well, Isolate the $t$. $\sqrt t =\frac {\sqrt 2}2$. Now how can we undo the square root? $\endgroup$ – lulu Oct 8 '16 at 14:56
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In the question the variables t and x are the same thing so x=t In order for us to find the right value of x that makes the function equal 3√2

We will solve the equation 4√x + √2 = 3√2

Implies : 4√x = 2√2

Implies : 2√x = √2

Implies : √x = (√2)/2

Implies :(√x)²= ((√2)/2)²

Hence : x = 1/2

Computing f(1/2) :

f(1/2) = 4√(1/2) + √2

f(1/2) = 4/√2 + √2

f(1/2) = (4+2)/√2

f(1/2) = 6/√2

f(1/2) = (6√2)/2

f(1/2) = 3√2

Hope you find this helpful.

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  • $\begingroup$ Thanks this was very helpful :) $\endgroup$ – Jude Molloy Oct 8 '16 at 15:12
  • $\begingroup$ I'm very touched :) $\endgroup$ – FuzzyPixelz Oct 8 '16 at 15:13
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    $\begingroup$ Nicely answered. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 9 '16 at 10:33
  • $\begingroup$ Thank You ! I will definitely learn it ! $\endgroup$ – FuzzyPixelz Oct 9 '16 at 10:41
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You have $f(x)$.You have to find a suitable $f(t)$.This means when you plug in $t$ for every value of $x$ so that the result becomes $3\sqrt2$.

If $f(x)=4\sqrt x+\sqrt 2$ then $f(t)=4\sqrt t+\sqrt 2$.

Now,solve. $$4\sqrt t+\sqrt2=3\sqrt2$$

$$4\sqrt t=2\sqrt2$$

$$\sqrt t=\frac {1}{\sqrt2}$$

$$t=\frac 12$$.

You can back-calculate to check $f(\frac12)=3\sqrt2$

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  • $\begingroup$ Since the op is trying to solve for $t$, you should probably replace every x by to, so that $f(t) = 4\sqrt t + \sqrt 2 = 3\sqrt 2$, so in the end, $t = \frac 12$ $\endgroup$ – Namaste Oct 8 '16 at 15:05
  • $\begingroup$ @amWhy I'm writing this from my mobile lying in the hospital bed down with fever....not in a position to edit it...thanks for your suggestion though...will try to do it once I am back home.... $\endgroup$ – tatan Oct 9 '16 at 5:37

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