Algebra Chapter 3 Quadratic Surds
⇔ If n is a positive integer and fl in a positive rational number, which can not express the nth power of some rational number, then the irrational number \(\sqrt[n]{a} \text { or } a^{\frac{1}{n}}\) that is the positive nth root of a is called surdor a radical.
⇒ The symbol \(\sqrt[n]{ }\) is called the radical sign n is the called the radical sign, n is called the order of the surd (or radical) and a is called the radicand.
⇒ Hence \(\sqrt[3]{\sqrt{2}}\) is not a surd as 2 is not a rational number.
⇒ A surd of order 2 is called a quadratic surd or square root.
⇒ Hence \(\sqrt{7}, \sqrt{\frac{4}{7}}\) are quadratic surd.
- If a is only a positive rational number that is not the square of a rational number, then ±√a type of number is called pure quadratic surd.
- Again a number of the form a±√b or ±a√b is called mixed quadratic surds where a is a rational number (≠0) and √b is a pure quadratic surd.
- Two or more quadratic surds are said to be similar surds if they can be expressed as rational multiple of the same surd.
- If a and b are such numbers prime Jo each other (i.e. HCF of a and b is 1) and neither of which is perfect square, then √a, √b are dissimilar surds.
- Any factor multiplying with any surds, if the product is free from surds then the factor, is called (the Rationalisation factor of that surround the process is called the Rationalisation.
- If the sum and the product of any mixed quadratic surd with rationalising factor are both rational number, the mixed quadratic surd is said to be conjugate or complementary surd.
Read and Learn More WBBSE Solutions for Class 10 Maths
Maths Algebra Chapter 3 Quadratic Surds True Or False
Example 1. √75 and √147 are similar surd.
Solution: True
Example 2. √π is a quadratic surd.
Solution: False
Example 3. Product of two surds is a surd.
Solution: False
Class 10 Maths Algebra Chapter 3 Solutions
Example 4. Product of two surds is not a surd.
Solution: False
Example 5. Sum of two surds is a surd.
Solution: False
Example 6. √6 + √3 = √7+ √7 .
Solution: False
Example 7. √6+ √ 3 > √7 + √2
Solution: False
Example 8. π is an irrational number which is not a surd.
Solution: True
Factorization Class 10 Solutions
Example 9. √32 is a surd but \(\sqrt[5]{32}\) is not a surd.
Solution: False
Example 10. Conjugate surd of √5-2 is √5 + 2
Solution: False
Maths Algebra Chapter 3 Quadratic Surds Fill In The Blanks
Example 1. 5√11 a _______ number (Rational/irrational)
Solution: Irrational
Example 2. Conjugate surd of √3-5 is _______
Solution: -√3 – √5
Example 3. If the product and sum of two quadratic surd is a rational number, then surds are ________ surd.
Solution: Conjugate
Class 10 Algebra Chapter 3 Solved Examples
Example 4. Rationalising factor of 7- √3 is _______
Solution: 7 + √3 or, -7 – √3
Example 5. Product of \(3^{\frac{1}{2}}\) and √3 is _______
Solution: 3
Example 6. √6 x √15 = x√10 then x = ________
Solution: 3
Example 7. Conjugate of -√5-1 is________
Solution: √5 -1
Wbbse Class 10 Algebra Notes
Example 8. To denote square root, we use √ instead of ______
Solution: \(\sqrt[2]{ }\)
Example 9. √108 – √75 = ______
Solution: √3
Example 10. Two quadratic surd √8 and √32 are rational multiple of same surd _______
Solution: √2
Maths Algebra Chapter 3 Quadratic Surds Short Answer Type Questions
Example 1. x = 3 + 2√2 let us write the value of x + \(\frac{1}{x}\)
Solution: \(\frac{1}{x}=\frac{1}{3+2 \sqrt{2}}\)
= \(\frac{3-2 \sqrt{2}}{(3+2 \sqrt{2})(3-2 \sqrt{2})}=\frac{(3-2 \sqrt{2}}{9-8}=3-2 \sqrt{2}\)
x + \(\frac{1}{x}\) = 3 + 2√2 + 3 -2√2 = 6
∴ The value of x + \(\frac{1}{x}\) = 6.
Example 2. Which is greater of (√15+√3) and (√10 + √8)
Solution: √15 + √10 > √8 + √3
⇒ or, \(\frac{1}{\sqrt{15}+\sqrt{10}}<\frac{1}{\sqrt{8}+\sqrt{3}}\),
⇒ or, \(\frac{\sqrt{15}-\sqrt{10}}{15-10}<\frac{\sqrt{8}-\sqrt{3}}{8-3}\)
⇒ or, √15 – √10 < √8 – √3
⇒ or, √15 + √3 < √10 + √8
∴ The value of x + \(\frac{1}{x}\) = 6.
Example 3. Let us write what should be subtracted from √72 to get √32.
Solution: √72 -√32 = \(\sqrt{4 \times 2 \times 9}-\sqrt{16 \times 2}\)
= 6√2 -4√2 = 2√2
Factorization Formulas Class 10
Example 4. Simplify \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)
Solution: Rationalising, we get
⇒ \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\)
= \(\sqrt{2}-1+\sqrt{8}-\sqrt{2}+2-\sqrt{3}=1\)
Example 5. \(\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}=?\)
Solution: \(\frac{\sqrt{4 \times 6}+\sqrt{6 \times 6 \times 6}}{\sqrt{6 \times 16}}\)
= \(\frac{2 \sqrt{6}+6 \sqrt{6}}{4 \sqrt{6}}=\frac{8 \sqrt{6}}{4 \sqrt{6}}=2\)
Example 6. 2x = √5 +1, x2 – x – 1 = ?
Solution: x2 – x- 1
= \(\left(\frac{\sqrt{5}+1}{2}\right)^2-\left(\frac{\sqrt{5}+1}{2}\right)-1=\frac{5+1+2 \sqrt{5}}{4}-\frac{\sqrt{5}+1}{2}-1\)
= \(\frac{6+2 \sqrt{5}-2 \sqrt{5}-2-4}{4}=\frac{0}{4}=0\)
Example 7. \(x=\sqrt{3}+\sqrt{2}, \quad x^2+\frac{1}{x^2}=?\)
Solution: \(\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{3-2}\)
= \(\sqrt{3}-\sqrt{2}\)
\(\left(x^2+\frac{1}{x^2}\right)=\left(x+\frac{1}{x}\right)^2-2 x \frac{1}{x}\)
= \((\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})^2-2=4 \times 3-2=10\)
Example 8. x + y = √3, = √2, 8xy (x2+y2) = ?
Solution: 8xy (x2 + y2)
⇒ x + y ≠ x- y = √3+√2
= 4xy x 2 (x2 + y2)
= [{(x + y)2– (x – y)2} {(x + y)2} + {(x – y)2} (x + y)- x – y] = √3 – √2
={(√3)2-(√2)2} {(√3)2+(√2)2}
= (3 – 2) (3 + 2) = 5
Class 10 Maths Algebra Important Questions
Example 9. \(x+\frac{1}{x}=\sqrt{3} ; \quad x^3+\frac{1}{x^3}=?\)
Solution: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3 x \frac{1}{x}\left(x+\frac{1}{x}\right)\)
= \((\sqrt{3})^3-3 \sqrt{3}=3 \sqrt{3}-3 \sqrt{3}=0\)
Example 10. If \(\frac{\sqrt{7}-2}{\sqrt{7}+2}=a \sqrt{7}+b\) then the value of b.
Solution: \(\frac{(\sqrt{7+2})(\sqrt{7}-2)}{7-4}=a \sqrt{7}+b\)
⇒ or, \(\frac{7+4-4 \sqrt{7}}{3}=a \sqrt{7}+b\)
⇒ or, \(\frac{11}{3}-\frac{4}{3} \sqrt{7}=a \sqrt{7}+b\) comparing relation and irrational part,
⇒ a = \(-\frac{4}{3}, b=\frac{11}{3}\).