Algebra Examples

$f (x) = \sqrt{2 x}$ , $g (x) = \sqrt{32 x} - \sqrt{50}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\sqrt{32 x} - \sqrt{50})$ by substituting in the value of $g$ into $f$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (\sqrt{32 x} - \sqrt{50})}$

Step 3

Rewrite $32 x$ as ${(2^{2})}^{2} \cdot (2 x)$ .

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Step 3.1

Factor $16$ out of $32$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (\sqrt{16 (2) x} - \sqrt{50})}$

Step 3.2

Rewrite $16$ as $4^{2}$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (\sqrt{4^{2} \cdot (2 x)} - \sqrt{50})}$

Step 3.3

Rewrite $4$ as $2^{2}$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (\sqrt{{(2^{2})}^{2} \cdot (2 x)} - \sqrt{50})}$

Step 3.4

Add parentheses.

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (\sqrt{{(2^{2})}^{2} \cdot (2 x)} - \sqrt{50})}$

Step 4

Pull terms out from under the radical.

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (2^{2} \sqrt{2 x} - \sqrt{50})}$

Step 5

Raise $2$ to the power of $2$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (4 \sqrt{2 x} - \sqrt{50})}$

Step 6

Rewrite $50$ as $5^{2} \cdot 2$ .

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Step 6.1

Factor $25$ out of $50$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (4 \sqrt{2 x} - \sqrt{25 (2)})}$

Step 6.2

Rewrite $25$ as $5^{2}$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (4 \sqrt{2 x} - \sqrt{5^{2} \cdot 2})}$

Step 7

Pull terms out from under the radical.

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (4 \sqrt{2 x} - (5 \sqrt{2}))}$

Step 8

Multiply $5$ by $- 1$ .

$f (\sqrt{32 x} - \sqrt{50}) = \sqrt{2 (4 \sqrt{2 x} - 5 \sqrt{2})}$