Evaluate 5 square root of 24*(2 square root of 28)

$5 \sqrt{24} \cdot (2 \sqrt{28})$

Step 1

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$5 \sqrt{4 (6)} \cdot (2 \sqrt{28})$

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

$5 \sqrt{2^{2} \cdot 6} \cdot (2 \sqrt{28})$

$5 \sqrt{2^{2} \cdot 6} \cdot (2 \sqrt{28})$

Step 2

Pull terms out from under the radical.

$5 (2 \sqrt{6}) \cdot (2 \sqrt{28})$

Step 3

Multiply $2$ by $5$ .

$10 \sqrt{6} \cdot (2 \sqrt{28})$

Step 4

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$10 \sqrt{6} \cdot (2 \sqrt{4 (7)})$

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

$10 \sqrt{6} \cdot (2 \sqrt{2^{2} \cdot 7})$

$10 \sqrt{6} \cdot (2 \sqrt{2^{2} \cdot 7})$

Step 5

Pull terms out from under the radical.

$10 \sqrt{6} \cdot (2 (2 \sqrt{7}))$

Step 6

Multiply $2$ by $2$ .

$10 \sqrt{6} \cdot (4 \sqrt{7})$

Step 7

Multiply $10 \sqrt{6} (4 \sqrt{7})$ .

Multiply $4$ by $10$ .

$40 \sqrt{6} \sqrt{7}$

Combine using the product rule for radicals.

$40 \sqrt{7 \cdot 6}$

Multiply $7$ by $6$ .

$40 \sqrt{42}$

$40 \sqrt{42}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$40 \sqrt{42}$

Decimal Form:

$259.22962793 \dots$

Related Questions