Calculate the value of $$\sqrt{80}+4\sqrt5$$ if $$2\sqrt5+\sqrt{125}=7x$$​​

Correct option is C

Given:

$$2\sqrt5+\sqrt{125}=7x$$

Formula Used:

$$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$​

Solution:

$$2\sqrt5+\sqrt{125}=7x$$

= $$2\sqrt5 + \sqrt{25 \times 5}$$​

= $$2\sqrt{5} + 5\sqrt{5}$$​​

= $$7\sqrt{5}$$

Then,

​$$7\sqrt{5} = 7x$$

x = $$\sqrt{5}$$

For $$\sqrt{80} + 4\sqrt{5}$$;

= $$\sqrt{16 \times 5} + 4\sqrt{5}$$​​

= $$4\sqrt{5} + 4\sqrt{5}$$

= $$8\sqrt{5}$$ 

= 8x   ( x = $$\sqrt5$$)​

Thus, the correct answer is (c).