Representing real numbers on the number line

Since (1^2=1) and (2^2=4), (\sqrt{3}) lies between (1) and (2). In exams, bracket square roots using perfect squares.

From (x+3=0), we get (x=-3). In exams, find the zero and show it as a point on the number line.

The zero of a polynomial is the value of \(x\) that makes the polynomial equal to zero. Thus, putting \(q(x)=0\), we get \(x-8=0\), so \(x=8\). Therefore, the zero is represented by the point 8 on the number line. Exam tip: the zero of a polynomial of the form \(x-a\) is \(a\), not \(-a\).

The zero (-5) is negative, so it lies to the left of (0). In exams, the sign of the zero tells its direction.

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(2x-6=0\Rightarrow 2x=6\Rightarrow x=3\). Therefore, the correct point on the number line is 3. The value 6 is the constant term, not the zero. Exam tip: Set a linear polynomial equal to zero and solve for \(x\) to find its zero.

The zero of a polynomial or function is the value of x for which its value becomes zero. Set r(x) equal to zero: 3x + 12 = 0. Subtract 12 from both sides to obtain 3x = −12, and divide by 3 to get x = −4. Therefore the zero is represented by the point −4 on the number line, so option B is correct. A common mistake is to report 4 by ignoring the negative sign, or to choose the constant term 12 or its negative without solving. Substitution verifies the result: 3(−4) + 12 = −12 + 12 = 0.

\(x=0\) represents the origin on the number line. Zero is neither positive nor negative; positive numbers lie to the right of \(0\), while negative numbers lie to its left. Exam tip: Never classify \(0\) as positive or negative.

Both (-1) and (1) are (1) unit away from (0). In exams, recognize opposite signs with equal distance.

The opposite point of (4) is (-4), at the same distance from (0) on the other side. In exams, only the sign changes for the opposite number.

The opposite point of (-7) is (7) because both are equally distant from (0). In exams, change the sign to identify the opposite number.

(\frac{5}{4}=1+\frac{1}{4}), so it is one-fourth after (1). In exams, convert an improper fraction into mixed form.

Since \(\frac{3}{10}=0.3\), we get \(2+\frac{3}{10}=2+0.3=2.3\). Option A, \(2.03\), represents \(2+\frac{3}{100}\), so it is incorrect. Exam tip: When the denominator is 10, the decimal form has one digit after the decimal point.

Numbers increase to the right on the number line, and (-1>-2). In exams, connect the right direction with the greater number.

Numbers decrease to the left on the number line, and (2.9<3). In exams, connect the left direction with the smaller number.

To compare closeness on a number line, calculate the absolute distance from the reference point 1. For 1/10, the distance is |1 − 1/10| = |9/10| = 9/10. For 9/10, the distance is |1 − 9/10| = |1/10| = 1/10. Since 1/10 is less than 9/10, the point 9/10 is closer to 1. Thus option B is correct. Option A reverses the distances, while option C would be true only if the two distances were equal. Option D is impossible because both fractions identify valid points on the number line.

(-0.2) is negative and greater than (-1). In exams, place small negative decimals between (-1) and (0).

The number 0.05 is positive, so it is greater than 0. It is also less than 1; therefore, it lies between 0 and 1. Option D is incorrect because 0.05 must not be confused with 5. Exam tip: First look at the integer part of a decimal to identify the interval between consecutive integers.

The distance between two points on a number line is the absolute value of the difference of their coordinates. Thus, distance = |1 − (−3)| = |4| = 4 units. The answer is not 2 units because the points lie four unit intervals apart. In exams, use brackets when subtracting a negative number.

The distance between two points on a number line is the absolute difference of their coordinates. Thus, the distance is \(|4-2.5|=1.5\) units. Option C is only the greater coordinate, while option D is the sum of the coordinates. In an exam, use the absolute difference for distance so that the answer is positive.

For (p(x)=x), putting (x=0) gives (p(x)=0). In exams, a zero is the value where the polynomial becomes (0).