What is the square of \(\sqrt{2+\sqrt{3}}\) equal to?
The square of a square root gives the number inside. So \(\left(\sqrt{2+\sqrt{3}}\right)^2=2+\sqrt{3}\).
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SubjectsMathematics
अपरिमेय संख्याएँ
In this Class 9 Mathematics topic from the Number Systems chapter, students learn that irrational numbers cannot be written in the form p/q, where p and q are integers and q is not zero. They explore familiar examples such as √2 and π, understand their non-terminating, non-repeating decimal expansions, and distinguish them from rational numbers. The topic also develops skills for representing irrational numbers on the number line and understanding their place within the real number system.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
The square of a square root gives the number inside. So \(\left(\sqrt{2+\sqrt{3}}\right)^2=2+\sqrt{3}\).
In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no repeating block is formed. Its decimal expansion is non-terminating and non-repeating; hence it is irrational. \(0.\overline{27}\) is rational because it repeats. Exam tip: non-terminating, non-repeating decimals are irrational.
If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence \(x+r\) is irrational. Exam tip: for \(r=0\), \(xr=0\), so a product is not always irrational.
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational, so A disproves the claim. In B, the product is \(\sqrt{6}\), still irrational. Exam tip: combine radicals before deciding the type of number.
If the sum of an irrational and a rational number were rational, subtracting the same rational number would make the irrational number rational, which is impossible. Exam tip: test “always” claims with counterexamples.
Multiplying by the conjugate gives denominator (9-2=7) and numerator ((3+\sqrt{2})^2=11+6\sqrt{2}). So the correct form is (\frac{11+6\sqrt{2}}{7}).
The product of irrational numbers need not be irrational. Here, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), a rational number. Exam tip: combine square roots first before deciding the type of number.
(\sqrt{108}=6\sqrt{3}), (\sqrt{48}=4\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). Therefore the result is (4\sqrt{3}).
The direct answer is A: \(\frac{\sqrt7-2}{3}\). To remove the radical from the denominator, multiply numerator and denominator by the conjugate of \(\sqrt7+2\), namely \(\sqrt7-2\). Thus \(p=\frac1{\sqrt7+2}\times\frac{\sqrt7-2}{\sqrt7-2}=\frac{\sqrt7-2}{(\sqrt7+2)(\sqrt7-2)}\). Using \((a+b)(a-b)=a^2-b^2\), the denominator is \(7-4=3\). Hence option A is correct. Option B, \(\sqrt7-2\), misses the denominator 3. Option C keeps the original sign and does not rationalise the denominator; it is not equal to the original fraction. Option D, \(2-\sqrt7\), is the negative of the numerator and therefore has the wrong sign. The conjugate changes the middle signs, and multiplying conjugates produces a difference of squares.
The direct answer is A: \(5+\sqrt6\). Let \(x=\sqrt{5+\sqrt6}\). The expression is \(x\times x=x^2\). By the definition of a square root, the square of the principal square root of a non-negative number is the number inside it. Since \(5+\sqrt6>0\), the value is exactly \(5+\sqrt6\). Option A is correct. Option B, \(25+6=31\), incorrectly squares the two parts separately; the expression is not \((5+\sqrt6)^2\). Option C, \(\sqrt{11}\), has no valid rule behind it and does not represent the number inside the radical. Option D, \(5-\sqrt6\), changes the plus sign without reason. A common mistake is to think that multiplying two identical radicals requires expanding the inside. It does not: \(\sqrt a\times\sqrt a=a\) for \(a\ge0\).
The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no fixed repeating block occurs. A rational number has a terminating or recurring decimal expansion. Exam tip: check repetition, not merely the digits used.
Multiplying by the conjugate gives (\frac{(\sqrt{3}+1)^2}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}). Rationalise the denominator.
If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. But \(x^2\) need not be irrational; \((\sqrt{2})^2=2\). Exam tip: adding a rational number preserves irrationality.
Since (8+\sqrt{15}) is positive, its square root is real. Squaring it gives the inside number (8+\sqrt{15}).
Multiplying by the conjugate makes the denominator (10-6=4). So (\frac{4(\sqrt{10}+\sqrt{6})}{4}=\sqrt{10}+\sqrt{6}).
Option A is correct: the value is 125. First simplify the radicals: \(\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5\) and \(\sqrt{45}=\sqrt{9\cdot5}=3\sqrt5\). Their sum is \(2\sqrt5+3\sqrt5=5\sqrt5\). Now square it: \((5\sqrt5)^2=25\times5=125\). Option A matches. Option B, 65, would result from an incorrect expansion or from mishandling the cross term. Option C, \(25\sqrt5\), stops after squaring the coefficient and forgets that \((\sqrt5)^2=5\); it is not the value of the full square. Option D, 100, may come from treating \(\sqrt{20}+\sqrt{45}\) incorrectly or ignoring the radical contribution. A safe method is to simplify each radical before adding like surds, then square. The memory cue is: extract perfect-square factors first; after combining, square both the number and the radical.
Both \(\sqrt{18}\) and \(\sqrt{2}\) are irrational, but \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), which is rational. Thus the claim is false. Exam tip: one counterexample disproves an “always” statement.
\(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the bracket is \(2\sqrt{3}\). Multiplying by \(\sqrt{3}\) gives (6).
Multiplying by the conjugate makes the denominator (8-5=3). So (\frac{3(\sqrt{8}-\sqrt{5})}{3}=\sqrt{8}-\sqrt{5}).
The sum of a rational number and an irrational number is always irrational. If their sum were rational, subtracting the rational number would make the irrational number rational, which is impossible. Exam tip: remember this rule for both addition and subtraction.
(\sqrt{162}=9\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{50}=5\sqrt{2}). Therefore the result is (11\sqrt{2}).
(\sqrt{48}=4\sqrt{3}) and (\sqrt{75}=5\sqrt{3}), so (s=9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).
Multiplying by the conjugate gives (\frac{(\sqrt{5}-\sqrt{2})^2}{5-2}). So the answer is (\frac{7-2\sqrt{10}}{3}).
The square root of a positive integer is rational only when that integer is a perfect square. Hence, if \(n\) is not a perfect square, \(\sqrt{n}\) is irrational. For example, \(12\) is not a perfect square, so \(\sqrt{12}\) is irrational. Exam tip: being even or odd alone is not enough.
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Since both factors are irrational, this is a counterexample to Reema’s “always” claim. Exam tip: combine square roots first before deciding the type of number.
QUIZ COMPLETE