Question

question 41 of 41
which of the following is equivalent to the radical expression below when ( x geq 0 )?
( sqrt{4x} \bullet sqrt{x + 7} )

a. ( sqrt{x^2 + 7x} )
b. ( sqrt{x^2 + 28x} )
c. ( sqrt{4x^2 + 28x} )
d. ( sqrt{4x^2 + x} )

Explanation:

Step1: Simplify \(\sqrt{4x}\)

We know that \(\sqrt{4x}=\sqrt{4}\cdot\sqrt{x}=2\sqrt{x}\) (since \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) for \(a\geq0,b\geq0\) and \(\sqrt{4} = 2\)). But also, we can use the property of radicals \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (for \(a\geq0,b\geq0\)) directly on the original expression \(\sqrt{4x}\cdot\sqrt{x + 7}\).

Step2: Apply the radical multiplication property

Using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (where \(a = 4x\) and \(b=x + 7\) and since \(x\geq0\), \(4x\geq0\) and \(x + 7\geq7>0\)), we have:

$$ \sqrt{4x}\cdot\sqrt{x + 7}=\sqrt{(4x)\cdot(x + 7)} $$

Step3: Expand the product inside the radical

Now, expand \((4x)\cdot(x + 7)\):

$$ (4x)\cdot(x + 7)=4x\cdot x+4x\cdot7 = 4x^{2}+28x $$

So, \(\sqrt{(4x)\cdot(x + 7)}=\sqrt{4x^{2}+28x}\)

Answer:

C. \(\sqrt{4x^{2}+28x}\)