Question

simplify (3+root 7) (2+root5)

Answer 1

$(3 + \sqrt{7} )(2 + \sqrt{5} ) \\ = > 3(2 + \sqrt{5} ) + \sqrt{7} (2 + \sqrt{5} ) \\ = > 6 + 3 \sqrt{5} + 2 \sqrt{7} + \sqrt{35}$

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Answer 2

Concept: All real numbers that are not rational numbers are referred to be irrational numbers in mathematics. In other words, it is impossible to describe an irrational number as the ratio of two integers.

• Decimals with non-terminating and non-repeating patterns make up irrational numbers.

• Only real numbers are used here.

• When one rational number and one irrational number is added together, only an irrational number remains as the sum. When one rational numbers, x, and an irrational number, y, are added together, the result is an irrational number, i.e., $x + \sqrt{y}$

• Any irrational number will produce an irrational number as a product when multiplied by any nonzero rational number. When a rational number y is multiplied to an irrational number x, the result is also irrational.

• There could or might not be a least common multiple (LCM) between any two irrational numbers.

• Two irrational numbers added together, subtracted from, multiplied, and divided may or may not result in a rational number.

Given: $(3 + \sqrt{7}) (2 + \sqrt{5} )$

To find: the product of the given irrational numbers

Solution:

We will multiply both the irrational numbers using the distributive property,

$(3 + \sqrt{7}) (2 + \sqrt{5} )$ =

$(3 * 2) + (3 * \sqrt{5} ) + (\sqrt{7} * 2) + (\sqrt{7} * \sqrt{5} )\\\\6 + 3\sqrt{5} + 2\sqrt{7} + \sqrt{35}$

Therefore, the product of $(3 + \sqrt{7}) (2 + \sqrt{5} )$ is $6 + 3\sqrt{5} + 2\sqrt{7} + \sqrt{35}$.

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