Problems from Product, division and sum of square roots

Calculate the square roots of the following perfect squares:

  1. $$\sqrt{1521}$$. Notice that $$1521=9\cdot169=(3\cdot3)\cdot(13\cdot13)$$, therefore we are sure that it is a perfect square.

  2. $$\sqrt{\dfrac{2916}{484}}$$. Notice that $$2916=36\cdot81$$ and $$494=121\cdot4$$.

  3. $$\sqrt{64}+3\sqrt{81}$$

  4. $$3\sqrt{19}-2\sqrt{49}+\dfrac{\sqrt{475}}{5}$$ where $$475=19\cdot25$$
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Development:

  1. As is indicated, $$1521=9\cdot169$$ so we can write the root as $$\sqrt{1521}=\sqrt{9\cdot169}$$.

Now, using the first property of the square root, we have: $$$\sqrt{1521}=\sqrt{9\cdot169}=\sqrt{9}\cdot\sqrt{169}=3\cdot13=39$$$

  1. Writing inside the root the information given in the statement, we have $$\sqrt{\dfrac{2916}{484}}=\sqrt{\dfrac{36\cdot81}{121\cdot4}}$$ and if we use the second property of the square root $$$\sqrt{\dfrac{36\cdot81}{121\cdot4}}=\dfrac{\sqrt{36\cdot81}}{\sqrt{121\cdot4}}$$$ Now we proceed as in the previous paragraph, using the first property of roots and: $$$\dfrac{\sqrt{36\cdot81}}{\sqrt{121\cdot4}}=\dfrac{\sqrt{36}\sqrt{81}}{\sqrt{121}\sqrt{4}}=\dfrac{6\cdot9}{11\cdot4}=\dfrac{54}{44}$$$

  2. With the table of most common perfect squares we have $$$\sqrt{64}+3\sqrt{81}=8+3\cdot9=8+27=35$$$

  3. We realise that we cannot calculate the root of $$19$$ because it is not a perfect square, therefore we just solve the other roots and then gather together the terms with a root on the one hand, and those that do not on the other $$$3\sqrt{19}-2\sqrt{49}+\dfrac{\sqrt{475}}{5}=3\sqrt{19}-2\cdot7+\dfrac{\sqrt{25\cdot19}}{5}=$$$ $$$=3\sqrt{19}-2\cdot7+\dfrac{\sqrt{25}\cdot\sqrt{19}}{5}=3\sqrt{19}-14+\dfrac{5\cdot\sqrt{19}}{5}=$$$ $$$=(3+\dfrac{5}{5})\sqrt{19}-14=4\sqrt{19}-14$$$

Solution:

  1. $$39$$
  2. $$\dfrac{54}{44}$$
  3. $$35$$
  4. $$4\sqrt{19}-14$$
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