Find $x$ such that $4\sqrt x+\sqrt 2=3\sqrt 2$

Question

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 $$f(t)=3\sqrt 2$$

Answer 1

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.

Answer 2

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 $f(\frac12)=3\sqrt2$