Representing real numbers on the number line

The governing concept is simplifying a square root by extracting perfect-square factors. Factor 50 as 25 × 2, where 25 is a perfect square. Using √(ab) = √a × √b, we get √50 = √(25 × 2) = √25 × √2 = 5√2. Therefore option A is correct. A numerical check gives √50 approximately 7.07, and 5√2 has the same approximate value. Although 2√5 is also approximately 4.47 and is not equal to √50, the other choices are much larger: 10√5 and 25√2 do not preserve the original value. The simplified radical should have no square factor remaining inside the root, which is why 5√2 is the preferred form.

(3\sqrt{2}) is about (4.24), so it lies between (4) and (5). In exams, you may estimate (\sqrt{2}) as about (1.414).

The governing concept is that the greater real number lies farther to the right on a number line. Compare the values without relying only on rounded decimals. Since 3.5 is positive, square it: 3.5²=12.25. Because 12<12.25 and both sides are non-negative, taking square roots preserves the order, so √12<√12.25=3.5. Numerically, √12 is approximately 3.464, which confirms the exact comparison. Therefore b=3.5 is to the right, making option B correct. Option A chooses the smaller value, option C is false because the values are unequal, and option D is unnecessary because the comparison is fully determined by squaring positive quantities.

(\frac{19}{6}=3+\frac{1}{6}), so it lies between (3) and (4). In exams, divide to identify the whole part.

(-\frac{22}{7}) is about (-3.14), so it lies between (-4) and (-3). In exams, note the order of negative decimals.

When the interval is divided into 10 equal parts, each part has a value of \\(\frac{1}{10}\\). Since \\(0.7=\frac{7}{10}\\), the seventh mark from 0 represents 0.7. The eighth mark represents \\(0.8\\), so it is incorrect. In exams, convert the decimal into a fraction with denominator 10, 100, and so on to identify the mark.

The exact midpoint of two numbers is their average: \(\frac{1.8+2.6}{2}=\frac{4.4}{2}=2.2\). Therefore, 2.2 is correct. The values 2.0 and 2.4 are not equally distant from both endpoints; for an exact midpoint, calculate the average of the two numbers.

There are two numbers at a distance of 1.5 units from 4 on the number line: \\(4-1.5=2.5\\) and \\(4+1.5=5.5\\). Since \\(x<4\\), the number to the left of 4 must be chosen; hence, \\(x=2.5\\). Exam tip: Always check the direction or inequality condition in distance questions.

Moving to the right on a number line means adding the given distance to the starting number. Thus, \\(y=-2+3.25=1.25\\). Option D results from subtracting the distance, which would represent moving to the left. Exam tip: add for a rightward move and subtract for a leftward move.

(-\frac{5}{4}=-1.25), which is to the left of (-1). In exams, convert negative fractions into decimals to check.

(-\frac{3}{2}=-1.5), so it lies between (-2) and (-1). In exams, the decimal form of a negative fraction helps.

From \(x^2-25=0\), we get \(x^2=25\). Therefore, \(x=5\) or \(x=-5\), so the solutions lie at −5 and 5 on the number line. Option B incorrectly includes 25; 25 is not a value of \(x\) satisfying the equation. Exam tip: When taking the square root of a positive number, consider both the positive and negative values.

From (2x-1=0), (x=\frac{1}{2}), and from (x+4=0), (x=-4). In exams, solve each linear factor separately.

(\sqrt{2}) is about (1.414), so the order is (0,\frac{1}{2},\sqrt{2},2). In exams, increasing order is from left to right.

(-\sqrt{2}) is about (-1.414) and (-\frac{3}{2}=-1.5), so (x) is greater. In exams, for negative numbers, the point with smaller distance from (0) may lie to the right.

Every real number corresponds to exactly one point on the number line. Irrational numbers such as \(\sqrt{2}\) have non-terminating, non-repeating decimals, so B and D are false. Exam tip: check whether an endless decimal repeats before classifying it.

Since \(5^2=25<26<36=6^2\), taking positive square roots gives \(5<\sqrt{26}<6\). Therefore, \(\sqrt{26}\) lies between 5 and 6. Exam tip: Compare the number with the two nearest consecutive perfect squares to locate its square root.

Each part is (\frac{3}{12}=\frac{1}{4}), so the seventh point is (\frac{7}{4}). Divide the total length by equal parts.

Since −√16 = −4, the distance is the absolute difference between the numbers: |2.5 − (−4)| = |6.5| = 6.5. Therefore, the correct answer is 6.5.

(\sqrt{6}) is about (2.45), so (-\sqrt{6}) is about (-2.45). (-2.4) is greater, so it is to the right.

The midpoint is the average of the two numbers: (−1.6 + 2.8) ÷ 2 = 1.2 ÷ 2 = 0.6. Therefore, option B is correct.

(\sqrt{45}) is about (6.7), so it is closer to (7). Nearby perfect squares (36) and (49) help in estimation.

Moving left on the number line means subtracting the distance from the starting coordinate. Thus, the final coordinate is 1.25 - 2.75 = -1.5. Option C gives only the magnitude of the difference and ignores the negative direction. Exam tip: add for movement to the right and subtract for movement to the left.

(\frac{4}{3}\approx1.33) and (\sqrt{2}\approx1.41), so the order is (\frac{4}{3}<\sqrt{2}<1.5). Estimate values for comparison.

The direct answer is option C: between 7 and 8. To locate a square root, compare the number under the root with consecutive perfect squares. We have \(7^2=49\) and \(8^2=64\). Since \(49<50<64\), taking positive square roots gives \(7<\sqrt{50}<8\). Option A, 5 and 6, is wrong because their squares are 25 and 36, both below 50. Option B, 6 and 7, is wrong because 50 is greater than \(7^2=49\), so the root is slightly greater than 7. Option C is correct. Option D, 8 and 9, is wrong because \(50<64=8^2\), so the root is below 8. Memory trick: bracket a square root using the two neighbouring perfect squares.