If (p=\frac{1}{\sqrt{13}+3}), which is the simplified form of (p)?
Correct answer: A
Multiplying by the conjugate (\sqrt{13}-3) makes the denominator (13-9=4). So (p=\frac{\sqrt{13}-3}{4}).
Mathematics
Irrational numbers
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
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Hard · Level 5
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- (\frac{\sqrt{13}-3}{4})
- (\sqrt{13}-3)
- (\frac{\sqrt{13}+3}{4})
- (3-\sqrt{13})
Hard · Level 5
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- (9+\sqrt{20})
- (81+20)
- (\sqrt{29})
- (9-\sqrt{20})
Hard · Level 5
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- It is always rational.
- It is always irrational.
- It can be either rational or irrational.
- It must necessarily be zero.
Hard · Level 5
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- (9+4\sqrt{5})
- (9-4\sqrt{5})
- (1+\sqrt{5})
- (5+2\sqrt{5})
Hard · Level 5
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- (\frac{11+4\sqrt{6}}{5})
- (11+4\sqrt{6})
- (\frac{11-4\sqrt{6}}{5})
- (5)
Hard · Level 5
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- \(a+b\) is irrational
- \(a+b\) is rational
- \(a+b\) is always an integer
- The type of \(a+b\) depends on the value of \(b\)
Hard · Level 5
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- It is not real
- Its square is (10+\sqrt{21})
- It equals (10+\sqrt{21})
- It is a rational integer
Hard · Level 5
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- \(x+r\) is irrational
- \(x-r\) is rational
- \(rx\) is rational
- \(x/r\) is rational
Hard · Level 5
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- (\frac{2(\sqrt{15}+\sqrt{6})}{3})
- (\sqrt{15}+\sqrt{6})
- (6(\sqrt{15}+\sqrt{6}))
- (\frac{\sqrt{15}+\sqrt{6}}{6})
Hard · Level 5
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- (288)
- (148)
- (24\sqrt{2})
- (196)
Hard · Level 5
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- दोनों संख्याएँ बराबर हों
- एक संख्या दूसरी संख्या का योगात्मक प्रतिलोम हो
- दोनों संख्याएँ धनात्मक हों
- दोनों संख्याएँ अलग-अलग वर्गमूलों के रूप में हों
Hard · Level 5
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- (10)
- \(2\sqrt{5}\)
- (20)
- \(\sqrt{80}\)
Hard · Level 5
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- (\sqrt{12}-\sqrt{7})
- (5(\sqrt{12}-\sqrt{7}))
- (\frac{\sqrt{12}-\sqrt{7}}{5})
- (5\sqrt{84})
Hard · Level 5
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- The product of a non-zero rational number and an irrational number is irrational.
- The sum of two irrational numbers is always irrational.
- The difference between an irrational number and a rational number is always rational.
- The quotient of two irrational numbers is always irrational.
Hard · Level 5
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- (13\sqrt{2})
- (11\sqrt{2})
- (15\sqrt{2})
- (17\sqrt{2})
Hard · Level 5
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- (2\sqrt{13})
- (6)
- (2\sqrt{13}+6)
- (\sqrt{13})
Hard · Level 5
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- (13)
- (11)
- (9)
- (15)
Hard · Level 5
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- (\frac{5-\sqrt{21}}{2})
- (\frac{5+\sqrt{21}}{2})
- (\sqrt{7}-2)
- (1)
Hard · Level 5
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- The sum of any two irrational numbers is irrational.
- The product of any two irrational numbers is irrational.
- If \(r\) is a non-zero rational number and \(x\) is irrational, then \(r+x\) is irrational.
- Subtracting an irrational number from itself gives an irrational number.
Hard · Level 5
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- \(\sqrt{2},\ \sqrt{3},\ \sqrt{5}\)
- \(\sqrt{2},\ \sqrt{49},\ \sqrt{5}\)
- \(\sqrt{2}\times\sqrt{8},\ \sqrt{3},\ \sqrt{5}\)
- \(\frac{\sqrt{12}}{\sqrt{3}},\ \sqrt{7},\ \sqrt{11}\)
Hard · Level 5
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- (14)
- (25)
- (36)
- (10\sqrt{11})
Hard · Level 5
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- (12\sqrt{5})
- (10\sqrt{5})
- (8\sqrt{5})
- (\sqrt{360})
Hard · Level 5
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- (2\sqrt{8})
- (\sqrt{8})
- (2\sqrt{6})
- (7)
Hard · Level 5
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- \(\sqrt{2}+\sqrt{3}\)
- \(\sqrt{5}+(3-\sqrt{5})\)
- \(\sqrt{7}+\sqrt{7}\)
- \(\sqrt{2}+(1+\sqrt{3})\)
Hard · Level 5
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- \(\sqrt{7}+(-\sqrt{7})=0\)
- \(\sqrt{2}+\sqrt{3}\)
- \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\)
- \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\)
What is the value of (\sqrt{9+\sqrt{20}}\times\sqrt{9+\sqrt{20}})?
Correct answer: A
The direct answer is A, 9+√20. Let x = √(9+√20). The expression is x×x, and the basic rule is √a×√a = a whenever a is non-negative. Here 9+√20 is positive, so the product is exactly 9+√20. Option A is correct. Option B, 81+20 = 101, incorrectly squares the two parts separately and ignores the middle term in a square. Option C, √29, is not equal to 9+√20; it incorrectly combines the numbers under different radicals. Option D, 9−√20, is the conjugate-like expression and is not produced by multiplying the same square root by itself. No extra expansion is needed. A useful check is to call the whole quantity inside the radical one number rather than splitting it.
If two irrational numbers \(a\) and \(b\) are such that \(a+b\) is rational, which conclusion about \(ab\) is correct?
Correct answer: C
Option C is correct. For \(a=\sqrt{2}, b=-\sqrt{2}\), the sum is 0 and \(ab=-2\) is rational. But with \(b=1-\sqrt{2}\), the sum is 1 while \(ab=\sqrt{2}-2\) is irrational. Exam tip: verify a claim using contrasting examples.
What is the simplified form of (\frac{\sqrt{5}+2}{\sqrt{5}-2})?
Correct answer: A
The direct answer is A: \(9+4\sqrt5\). Rationalise the denominator by multiplying by its conjugate, \(\sqrt5+2\): \(\frac{\sqrt5+2}{\sqrt5-2}\times\frac{\sqrt5+2}{\sqrt5+2}=\frac{(\sqrt5+2)^2}{5-4}\). The denominator is 1, so the result is \((\sqrt5+2)^2\). Expanding gives \(5+4\sqrt5+4=9+4\sqrt5\). Option A is correct. Option B, \(9-4\sqrt5\), results from using the wrong sign in the expansion and is also the value of a different squared expression. Option C, \(1+\sqrt5\), does not follow from the square. Option D, \(5+2\sqrt5\), omits the final constant 4 and uses an incorrect middle coefficient. The safe method is conjugate first, then expand carefully.
If (x=\sqrt{8}+\sqrt{3}) and (y=\sqrt{8}-\sqrt{3}), what is the value of (\frac{x}{y})?
Correct answer: A
Multiplying by the denominator conjugate gives denominator (8-3=5). The numerator is ((\sqrt{8}+\sqrt{3})^2=11+4\sqrt{6}).
If \(a\) is an irrational number and \(b\) is a non-zero rational number, which statement about \(a+b\) is always true?
Correct answer: A
Assume \(a+b\) is rational. Then \(a=(a+b)-b\) would be the difference of two rational numbers and hence rational, a contradiction. Therefore \(a+b\) is irrational. Exam tip: rational ± irrational is always irrational.
Which statement is definitely true about (\sqrt{10+\sqrt{21}})?
Correct answer: B
Since (10+\sqrt{21}) is positive, its square root is real. Squaring it gives the inside number.
If \(x\) is an irrational number and \(r\ne 0\) is a rational number, which of the following statements is always true?
Correct answer: A
If \(x+r\) were rational, then \(x=(x+r)-r\) would be the difference of two rational numbers and hence rational, a contradiction. Thus A is correct. Exam tip: rational ± irrational is always irrational.
Which is the rationalised form of (\frac{6}{\sqrt{15}-\sqrt{6}})?
Correct answer: A
Multiplying by the conjugate makes the denominator (15-6=9). So (\frac{6(\sqrt{15}+\sqrt{6})}{9}=\frac{2(\sqrt{15}+\sqrt{6})}{3}).
What is the value of ((\sqrt{50}+\sqrt{98})^2)?
Correct answer: A
The direct answer is option A, 288. Simplify each radical first: √50 = √(25 × 2) = 5√2, and √98 = √(49 × 2) = 7√2. Therefore √50 + √98 = 5√2 + 7√2 = 12√2. Now square the sum: (12√2)² = 12² × (√2)² = 144 × 2 = 288. Option A, 288, is correct. Option B, 148, is wrong because it does not result from correctly squaring 12√2. Option C, 24√2, is wrong because it is not the value of the square; it resembles an unsquared or incorrectly expanded result. Option D, 196, is wrong because it could arise from squaring 14, but the simplified sum is 12√2, not 14. One may also check by expansion: 50 + 98 + 2√(50×98) = 148 + 2√4900 = 148 + 140 = 288. The useful cue is to extract perfect-square factors before adding like radicals.
Which condition on two irrational numbers guarantees that their sum is rational?
Correct answer: B
If one irrational number is x and the other is its additive inverse −x, then \(x+(-x)=0\). Since 0 is rational, the sum is rational. Exam tip: for “guaranteed” statements, check whether every allowed case works, not just one example.
What is the value of \(\sqrt{5}\left(\sqrt{125}-\sqrt{45}\right)\)?
Correct answer: A
\(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so the bracket is \(2\sqrt{5}\). Multiplying by \(\sqrt{5}\) gives (10).
What is the rationalised form of (\frac{5}{\sqrt{12}+\sqrt{7}})?
Correct answer: A
Multiplying by the conjugate makes the denominator (12-7=5). So (\frac{5(\sqrt{12}-\sqrt{7})}{5}=\sqrt{12}-\sqrt{7}).
Which of the following statements correctly describes the relationship between rational and irrational numbers?
Correct answer: A
If \(q\ne0\) is rational and \(x\) is irrational, assuming \(qx\) rational gives \(x=(qx)/q\) rational, a contradiction. B fails because \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims with a counterexample.
Which is the simplified form of (\sqrt{242}+\sqrt{128}-\sqrt{72})?
Correct answer: A
(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). Therefore the result is (13\sqrt{2}).
If (t=\sqrt{13}+3), what is the value of (t+\frac{4}{t})?
Correct answer: A
The direct answer is option A, 2√13. Given t = √13 + 3, find 4/t by rationalising the denominator: 4/(√13 + 3) × (√13 − 3)/(√13 − 3). The denominator is 13 − 9 = 4, so the fraction becomes 4(√13 − 3)/4 = √13 − 3. Hence t + 4/t = (√13 + 3) + (√13 − 3) = 2√13. Option A is correct because the rationalised fraction cancels the +3 and −3 terms. Option B, 6, is wrong because the radical terms do not cancel; they add to 2√13. Option C, 2√13 + 6, is wrong because it incorrectly adds the constants instead of noticing that +3 and −3 cancel. Option D, √13, is wrong because there are two √13 terms, not one. The key idea is that √13 + 3 and √13 − 3 are conjugates, and their product is 13 − 9 = 4. This makes the fraction easy to simplify.
If (s=\sqrt{75}+\sqrt{192}), what is the value of (\frac{s}{\sqrt{3}})?
Correct answer: A
(\sqrt{75}=5\sqrt{3}) and (\sqrt{192}=8\sqrt{3}), so (s=13\sqrt{3}). Dividing by (\sqrt{3}) gives (13).
Which is the simplified form of (\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}})?
Correct answer: A
Multiplying by the conjugate gives (\frac{(\sqrt{7}-\sqrt{3})^2}{7-3}). So the answer is (\frac{5-\sqrt{21}}{2}).
A student has written the following conclusions about irrational numbers. Which conclusion is always correct?
Correct answer: C
Option C is correct. If \(r+x\) were rational, then \(x=(r+x)-r\) would also be rational, a contradiction. For A, \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims using a counterexample.
Which of the following options contains only irrational numbers?
Correct answer: A
Since 2, 3 and 5 are not perfect squares, \(\sqrt2\), \(\sqrt3\) and \(\sqrt5\) are all irrational. The other options contain \(\sqrt{49}=7\), \(\sqrt2\times\sqrt8=\sqrt{16}=4\), or \(\frac{\sqrt{12}}{\sqrt3}=\sqrt4=2\), which are rational. Exam tip: a square root of a perfect square is rational.
If a rectangle has length (5+\sqrt{11}) and breadth (5-\sqrt{11}), what will be its area?
Correct answer: A
Area is ((5+\sqrt{11})(5-\sqrt{11})=25-11=14). Multiplying conjugate dimensions can give a rational area.
What is the simplified form of (\sqrt{500}-\sqrt{320}+\sqrt{180})?
Correct answer: A
(\sqrt{500}=10\sqrt{5}), (\sqrt{320}=8\sqrt{5}), and (\sqrt{180}=6\sqrt{5}). So the result is (10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}).
What is the value of (\frac{1}{\sqrt{8}+\sqrt{6}}+\frac{1}{\sqrt{8}-\sqrt{6}})?
Correct answer: A
The first term becomes (\frac{\sqrt{8}-\sqrt{6}}{2}) and the second becomes (\frac{\sqrt{8}+\sqrt{6}}{2}). Their sum is (\sqrt{8}).
A student claims that the sum of two irrational numbers is always irrational. Which of the following expressions is a counterexample to the claim?
Correct answer: B
Both \(\sqrt{5}\) and \(3-\sqrt{5}\) are irrational, but their sum is \(\sqrt{5}+3-\sqrt{5}=3\), which is rational. Thus the claim is false. Exam tip: check whether irrational terms cancel.
A student claims that the sum of any two irrational numbers is always irrational. Which of the following examples proves the claim wrong?
Correct answer: A
Both \(\sqrt{7}\) and \(-\sqrt{7}\) are irrational, but they are additive inverses, so their sum is \(0\), a rational number. Thus the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.
QUIZ COMPLETE