Question

$8 \times \sqrt{242} - 5 \times \sqrt{50 } + 3 \times \sqrt{98}$
Explained Answer please ​

Answer 1

Answer:

$84 \sqrt{2}$

Step-by-step explanation:

The question asked is

We will do prime factorisation of all the radicals (roots) here.

$\sqrt{242}$$\sqrt{2 \times 11 \times 11}$

We will remove the pair of 11 and write 11 outside the radical

$11 \sqrt{2}$

$\sqrt{50}$

$\sqrt{2 \times 5 \times 5}$

We will remove the pair of 5 and write 5 outside the radical

$5 \sqrt{2}$

$\sqrt{98}$

$\sqrt{2 \times 7 \times 7}$

We will remove the pair of 7 and write 7 outside the radical

$7 \sqrt{2}$

Now we have

$8(11 \sqrt{2} ) - 5(5 \sqrt{2} ) + 3(7 \sqrt{2} )$

which is

$88 \sqrt{2 } - 25 \sqrt{2} + 21 \sqrt{2}$

Since all three numbers have the same radical, we can add them

$(88 - 25 + 21) \sqrt{2}$

which comes out as

$84 \sqrt{2}$

Hence the answer