Irrational numbers

The positive number is (\sqrt{19}) because ((\sqrt{19})^2=19). Since (19) is not a perfect square, it is irrational.

(\sqrt{2}) is irrational so its decimal is non-terminating and non-repeating. Do not conclude from only a few decimal digits.

Every real number is not rational because irrational numbers are also real. The number system includes both types.

(\sqrt{2}) and (\sqrt{5}) are both irrational because (2) and (5) are not perfect squares. Check each number separately.

The direct answer is A, \(5\sqrt{7}\). Simplify the first radical: \(\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\). Simplify the second: \(\sqrt{63}=\sqrt{9 imes7}=3\sqrt{7}\). Since both now have the same radical part, add their coefficients: \(2\sqrt{7}+3\sqrt{7}=5\sqrt{7}\). Option A is correct. Option B, \(7\sqrt{5}\), uses the wrong radicand and does not follow from either simplification. Option C, \(13\sqrt{7}\), adds 2 and 3 incorrectly; the coefficient sum is 5, not 13. Option D, \(\sqrt{91}\), incorrectly multiplies the numbers inside the roots; multiplication is not the operation here. The exam rule is to factor out perfect squares first and then combine only like radicals.

(\sqrt{72}=6\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the difference is (4\sqrt{2}). Simplify like radicals first.

The positive number is (\sqrt{37}) because ((\sqrt{37})^2=37). Since (37) is not a perfect square, it is irrational.

This decimal has no fixed repeating block. So it is a non-terminating non-repeating decimal and an irrational number.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the claim is false. In contrast, \(3\sqrt{2}\) remains irrational. Exam tip: test an “always” statement using one counterexample.

(\sqrt{44}=2\sqrt{11}) and (\sqrt{99}=3\sqrt{11}), so the sum is (5\sqrt{11}). Simplify before adding like radicals.

The governing concept is simplification and addition of like surds. A square factor can be taken outside a radical, so √12 = √(4×3) = 2√3. Substituting this into the expression gives x = 2√3 + 2√3. Because both terms contain the same irrational factor √3, their numerical coefficients may be added: 2 + 2 = 4. Therefore x = 4√3, making option D correct. Option A keeps only the first simplified term and ignores the second term. Option B uses an incorrect coefficient sum. Option C incorrectly treats the sum of two radical terms as one radical. Unlike ordinary multiplication, addition under radicals is not combined by adding radicands; only like surds can be combined directly.

The governing concept is reduction and subtraction of like surds. First extract the perfect-square factor from √8: √8 = √(4×2) = 2√2. Substituting this into the expression gives a = 2√2 − √2 = (2−1)√2 = √2. Therefore option B is correct. Option A adds the coefficients instead of subtracting them. Option C incorrectly combines the radicands 8 and 2 under one square root, even though subtraction of square roots does not work that way. Option D treats √2 as though it were an integer. Direct addition or subtraction is permitted only when the remaining radical parts are identical, as they are here.

The governing concept is simplifying a radical before performing subtraction. Since 12 = 4×3, √12 = √(4×3) = 2√3. Substituting this equivalent form into the expression gives x = 2√3 − 2√3. The two terms are identical and have opposite signs, so they cancel exactly: x = 0. Therefore option A is correct. Option B leaves one radical without justification, option C effectively adds the terms instead of subtracting them, and option D has no valid algebraic basis. The important step is not to treat √12 as an unrelated radical; it must first be rewritten using its perfect-square factor.

The governing concept is the square-root property √a×√a=a for a non-negative real number. Here the same quantity, √(4+√7), is multiplied by itself. Therefore the product is [√(4+√7)]²=4+√7. The radicand is positive because √7 is positive, so applying the property is valid. Hence option A is correct. Option B incorrectly squares the two parts separately and then adds them, even though (4+√7)² would include an additional mixed term. Option C is unrelated to the given expression, and option D changes the sign of the radical without justification. Recognising the identical square-root factors avoids unnecessary expansion.