Chapter 9 Quadratic surd Exercise 9.3
Question 1. If m + 1/m = √3, let us calculate the simplified value of (i) m2 + 1/m2 and (ii) m3 + 1/m3.
Solution (1): m2 + 1/m2 = (m + 1/m)2 – 2.m.1/m
=(√3)2-2
=3-2
= 1
Read and Learn More WBBSE Solutions For Class 10 Maths
Solution (2): m3 + 1/m3 = (m+1/m)3 -3.m.1/m(m+1/m)
=(√3)2-3√3
= 3√3-3√3
=0 Ans.
“WBBSE Class 10 Maths Quadratic Surd Exercise 9.3 solutions”
2. Let us show that√5+√3 / √5-√3 – √5 – √3 / √5+ √3 = 2√15.
Solution: √5+√3 / √5-√3 – √5 – √3 / √5+ √3 = 2√15
L.H.S.= √5+√3 / √5-√3 – √5-√3 / √5+√3.
= (√5+√3)²-(√5-√3)² / (√5-√3) (√5+√3)
= (5+3+2√15)-(5+3-2√15) / (√5)²-(√3)²
= 8+2√15-8+2√15 /5-3
=4√15 / 2
= 2√15 R.H.S.
Question 2.
1. √2 (2+ √3) / √3(√3+1) – √2 (2-√3) /√3(√3-1)
Solution : √2 (2+ √3) / √3(√3+1) – √2 (2-√3) /√3(√3-1)
= √2 / √3 [2+√3 / √3+1 – 2-√3 / √3-1]
= √2 (2√3 −2+3−√3)-(2√3+2−3−√3) / (√3)2-(1)2
= √2 / √3[2√3-2+3-√3-2√3-2+3+√3 / 3-1]
= √2 / √3 x 6-4/2
= √2 / √3 x 2/2
= √2 x √3 / √3.√3
=√6/3
√2 (2+ √3) / √3(√3+1) – √2 (2-√3) /√3(√3-1) =√6/3
2. 3√7 / √5+ √2 – 5√5 / √2 + √7 + 2√2 / √7 + √5
Solution: 3√7 / √5+ √2 – 5√5 / √2 + √7 + 2√2 / √7 + √5
= 3√/7(√5 – √2) / (√5+√2)(√5+√2) – 5√5(√7-√2) / (√7+√2)(√7-√2) + 2√5/(√7+√5)(√7-√5)
= (√5+√2) √5-√2) (√7 + √2) √7 – √2) + (√7+ √5 √7-√5)
= 3√7(√5-√2)/(√5)2-(√2)2 – 5√5(√7-√2) / (√7)2-(√2)2 + 2√2 / (√7)2-(√5)2
= 3√7(√5-√2) /5-2 – 5√5(√7-√2)/7-2 + 2√2(√7-√5)/7-5
= 3√2(√5-√2)/3 – 5√5(√7-√2)/5 + 2√2(√7-√5)/2
=√35-√14-√35+√10+√14-√10
= 0
3√7 / √5+ √2 – 5√5 / √2 + √7 + 2√2 / √7 + √5 = 0
“West Bengal Board Class 10 Maths Chapter 9 Quadratic Surd Exercise 9.3 solutions”
3. \(\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}\)
Solution:
\(\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}\)= \(\frac{4 \sqrt{3}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}-\frac{30(4 \sqrt{3}+\sqrt{18})}{(4 \sqrt{3}-\sqrt{18})(4 \sqrt{3}+\sqrt{18})}-\frac{\sqrt{18}(3+\sqrt{12})}{(3-\sqrt{12})(3+\sqrt{12})}\)
= \(\frac{4 \sqrt{3}(2+\sqrt{2})}{(2)^2(\sqrt{2})^2}-\frac{30(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3})^2-(\sqrt{18})^2}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3)^2-(\sqrt{12})^2}\)
= \(\frac{4 \sqrt{3}(2+\sqrt{2})}{4-2}-\frac{30(4 \sqrt{3}+\sqrt{2})}{16 \times 3-18}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{9-12}\)
= \(\frac{4 \sqrt{3}(2+\sqrt{2})}{2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{30}-\frac{3(3+2 \sqrt{3})}{-3}\)
= \(4 \sqrt{3}+2 \sqrt{6}-4 \sqrt{3}-3 \sqrt{2}+2 \sqrt{6}\)
= 4√6
4. \(\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
Solution:
\(\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}-\frac{4 \sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\)
= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}\)
= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{1}\)
= \(\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}+\sqrt{18}-\sqrt{12}\)
= 0
“WBBSE Class 10 Quadratic Surd Exercise 9.3 solutions explained”
Question 3. If x = 2, y = 3 and z = 6 let us write by calculating the value of \(\frac{3 \sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4 \sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)
Solution: \(\frac{3 \sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4 \sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)
= \(\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
= \(\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}+\sqrt{3})}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\)
= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}\)
= \(\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{1}\)
= \(\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}+\sqrt{18}-\sqrt{12}\)
= 0
“WBBSE Class 10 Maths Exercise 9.3 Quadratic Surd problem solutions”
Question 4. If x = √7+√6 let us calculate the simplified values of
1. x – 1/x
Solution: x-1/x = (√7+√6)+(√7-√6)
= √7+√6 – √7 +√6
=2√6
x-1/x =2√6
2. x + 1/x
Solution: x – 1/x
= (√7+√6) + (√7-√6)
=√7+√6+√7-√6
=2√7
x – 1/x = 2√7
3. x²+1/x²
Solution: x²+1/x²
=(x+1/x)² – 2.x.1/x
=(2√7)² – 2.1
= 28-2
= 26
x²+1/x² = 26
4. x³ +1/x³
Solution: x³+1/x³
= (x+1/x)³ – 3.x.1/x(x+1/x)
=(2√7)³-3.1.2√7
=8 x 7√7 – 6√7
= 50√7
x³+1/x³ = 50√7
Question 5. x+√x2-1/x-√x2-1+ x-√x2-1/x+√x2-1 If the simplified value is 14, let us write by calculating what is the value of x.
Solution: \(\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}\)
\(\frac{\left(x+\sqrt{x^2-1}\right)^2+\left(x-\sqrt{x^2-1}\right)}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\)= \(\frac{x^2+x^2-1+2 x \sqrt{x^2-1}+x^2+x^2-1-2 x \sqrt{x^2-1}}{(x)^2-\left(\sqrt{x^2-1}\right)^2}\)
= \(\frac{4 x^2-2}{x^2-\left(x^2-1\right)}=\frac{4 x^2-2}{x^2-x^2+1}=4 x^2-2\)
According to the problem,
4x2 – 2 = 14
Or, 4x2 = 14 + 2 = 16
x2 = \(\frac{16}{4}\) = 4
\(x= \pm \sqrt{4}= \pm 2\)
Question 6. If a= √5+1 / √5-1 and b= √5-1/√5+1, let us following expressions
\(a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^2+(\sqrt{5}-1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}\)= \(\frac{5+1+2 \sqrt{5}+5+1-2 \sqrt{5}}{5-1}=\frac{12}{4}=3\)
& \(a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^2-(\sqrt{5}-1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}\)
= \(\frac{(5+1+2 \sqrt{5})-(5+1-2 \sqrt{5})}{5-1}=\frac{5+1+2 \sqrt{5}-5-1+2 \sqrt{5}}{4}=\frac{4 \sqrt{5}}{4}=\sqrt{5}\)
ab = \(\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1\)
“Class 10 WBBSE Maths Exercise 9.3 Quadratic Surd step-by-step solutions”
1. \(\frac{a^2+a b+b}{a^2-a b+b^2}\)
Solution: \(\frac{a^2+a b+b}{a^2-a b+b^2}\)
= \(\frac{a^2+b^2+a b}{a^2+b^2-a b}\)
= \(\frac{(a+b)^2-2 a b+a b}{(a+b)^2-2 a b-a b}\)
= \(\frac{(a+b)^2-a b}{(a+b)^2-3 a b}\)
= \(\frac{(3)^2-1}{(3)^2-3.1}\)
= \(\frac{9-1}{9-3}\)
= \(\frac{8}{6}\)
= \(\frac{4}{3}\)
2. (a-b)³/(a+b)³
Solution: (a-b)³/(a+b)³
= (√5)³/(3)
=5√5/27
(a-b)³/(a+b)³ =5√5/27
3. \(\frac{3 a^2+5 a b+3 b^2}{3 a^2-5 a b+3 b^2}\)
Solution: \(\frac{3 a^2+5 a b+3 b^2}{3 a^2-5 a b+3 b^2}\)
= \(\frac{3 a^2+6 a b+3 b^2-a b}{3 a^2-6 a b+3 b^2+a b}\)
= \(\frac{3\left(a^2+2 a b+b^2\right)-1}{3\left(a^2-2 a b+b^2\right)+1}\)
= \(\frac{3(a+b)^2-a b}{3(a-b)^2+a b}\)
= \(\frac{3(3)^2-1}{3(\sqrt{5})^2+1}\)
= \(\frac{27-1}{3.5+1}\)
= \(\frac{26}{16}\)
= \(\frac{13}{8}\)
= \(1 \frac{5}{8}\)
4. \(\frac{a^3+b^3}{a^3-b^3}\)
Solution: \(\frac{a^3+b^3}{a^3-b^3}\)
= \(\frac{(a+b)^3-3 a b(a+b)}{(a-b)^3+3 a b(a-b)}\)
= \(\frac{(3)^3-3.1 \cdot 3}{(\sqrt{5})^3+3 \cdot 1 \cdot \sqrt{5}}\)
= \(\frac{27-9}{8 \sqrt{5}}\)
= \(\frac{18 \sqrt{5}}{8 \sqrt{5} \cdot \sqrt{5}}\)
= \(\frac{18 \sqrt{5}}{8 \times 5}\)
= \(\frac{9 \sqrt{5}}{20}\)
“WBBSE Class 10 Chapter 9 Quadratic Surd Exercise 9.3 solution guide”
Question 7. If x = 2 + √3, y = 2-√3, let us calculate the simplified value of:
Solution : x = 2+√3
∴ 1/x = 1/2-√3
= 2-√3/(2+√3)(2-√3)
=2-√3/4-3
=2-√3/1
2-√3
Again, y=2-√3
∴ 1/y = 1/2- √3
=(2+√3)/(2+√3)(2-√3)
=2+√3/4-3
=2+√3/1
2+√3
1. x-1/x
Solution: x-1/x
=(2+√3) – (2-√3)
= 2+√3-2+√3
=2√3
x-1/x =2√3
2. y²+1/y²
Solution: y²+1/y²
=(y+1/y)²-2.y.1/y
={(2-√3)+(2+√3)}²
=(2-√3+2+√3)²-2
={(4)²-2)}
= 16-2
= 14
y²+1/y² = 14
3. x³-1/x³
Solution: x³-1/x³
= (x-1/x)³+3.x.1/x(x-1/x)
=(2√3)³+3.1.2√3
=8.3.√3+6√3
=24√3+√3
=30√3
x³-1/x³ =30√3
4. xy+1/xy
Solution: xy+1/xy
=x.y
=(2+√3)(2-√3)
=(2)²-(√3)²
= 4-3
= 1
∴ xy +1/xy
1+1/1
=1+1
=2
xy+1/xy =2
“West Bengal Board Class 10 Maths Exercise 9.3 Quadratic Surd solutions”
5. 3x²-5xy+3y²
Solution: 3x²-5xy+3y²
=3x²-6xy+3y²+xy
=3(x²-2xy+y²)+xy
=3(x-y)²+xy
=3{(2+√3)-(2-√3)}²+1
=3 x 4 x3 +1
=36+1
=37
3x²-5xy+3y² =37
Question 8. If x = √7+√3/√7-√3 and xy=1, let us show that x²+xy+y²/x²-xy+y² = 12/11
Solution:
\(\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \& x y=1, show that \frac{x^2+x y+y^2}{x^2-x y+y^2}\)x = \(\frac{(\sqrt{7}+\sqrt{3})(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}=\frac{(\sqrt{7}+\sqrt{3})^2}{(\sqrt{7})^2(\sqrt{3})^2}=\frac{7+3+2 \cdot \sqrt{7} \cdot \sqrt{3}}{7-3}=\frac{10+2 \sqrt{21}}{4}\)
= \(\frac{2(5+\sqrt{21})}{4}=\frac{5+\sqrt{21}}{2}\)
As xy = 1
∴ \(y=\frac{1}{x}=\frac{2}{5+\sqrt{21}}=\frac{2(5-\sqrt{21})}{(5+\sqrt{21})(5-\sqrt{21})}\)
= \(\frac{2(5-\sqrt{21})}{(5)^2-(\sqrt{21})^2}=\frac{2(5-\sqrt{21})}{25-21}=\frac{2(5-\sqrt{21})}{4}=\frac{5-\sqrt{21}}{2}\)
x + y = \(\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=\frac{5+\sqrt{21}+5-\sqrt{21}}{2}=\frac{10}{2}=5\)
Now, \(\frac{x^2+x y+y^2}{x^2-x y+y^2}=\frac{x^2+y^2+x y}{x^2+y^2-x y}=\frac{x^2+y^2+x y}{x^2+y^2-x y}\)
= \(\frac{(x+y)^2-x y}{(x+y)^2-3 x y}=\frac{(5)^2-1}{(5)^2-3.1}=\frac{25-1}{25-3}=\frac{24}{22}=\frac{12}{11}\)
Chapter 9 Quadratic surd Exercise 9.3 Multiple Choice Question
Question 1. If x=2+√3, the value of x + 1/x is
1. 2
2. 2√3
3. 4
4. 2-√3
Solution:
x=2+√3
.. 1/x
=1/ 2+√3
=1x(2-√3)/(2+√3)(2-√3)
=2-√3/4-3
2-√3/1
2-√3
.. x+1/x = 2+v3+2-√3
=4
x+1/x =4
Answer. 3. 4
Question 2. If p + q = √13 and p-q= √5 then the value of pq is
1. 2
2. 18
3. 9
4. 8
Solution: We know,
pq = (p+q-p-q)/4
=(√13)2-(√5)2/4
= 13-5/4
=8/4
=2
The value of pq is 2
Answer. 1. 2
“Class 10 WBBSE Maths Exercise 9.3 solutions for Quadratic Surd”
Question 3. If a + b = √5 and a-b=√3, the value of (a² + b²) is
1. 8
2. 4
3. 2
4. 1
Solution: a²+ b²= (a+b)²+(a-b)2/2
= (√5)²+(√3)2/2
= 5+3/2
=8/2
=3
The value of (a²+ b²) is 3
Answer. 1. 8
Question 5. If we subtract √5 from √125, the value is
1. √80
2. √120
3. √100
4. none of these
Solution: √125-√5
= √5x5x5
= √5x5x5-√5-√5
=4√5
= √16×5
= 80
√125-√5 = 80
Answer. 1. √80
Question 6. The product of the bracketed terms (5 -√3), (√3 -1), (5+ √3), and (√3+1) is
1. 22
2. 44
3. 2
4. 11
Solution: (5-√3) (5+√3) (√3-1) (√3+1)
= {(5)2- (√3)2} {(√3)2- (1)}
=(23) x (3-1)
=22 x 2
= 44
(5-√3) (5+√3) (√3-1) (√3+1) = 44
Answer. 2. 44
Chapter 9 Quadratic surd Exercise 9.3 True Or False
1. √75 and √147 are similar surds.
Solution: √75 & √5x5x3
=5√3 & √147
= √7x7x3
=7√3
√75 & √5x5x3 =7√3
Answer. True
“WBBSE Class 10 Maths Quadratic Surd Exercise 9.3 answers”
2.√π is a quadratic surd.
Answer. False
Chapter 9 Quadratic surd Exercise 9.3 Fill In The Blanks
1. 5√11 is an Irrational number (rational/irrational)
2. Conjugate surd of (√3-5) is -√3-5.
3. If the product and sum of two quadratic surds is a rational number, then the surds are Irrational surd.
Chapter 9 Quadratic surd Exercise 9.3 Short Answers
Question 1. If x=3+2√2, let us write the value of x + 1/x
Solution: 1/x =1/3 +2√2
1x(3-2√2) / (3+2√2) (3-2√2)
=3-2√2/9-8
=3-2√2
=3+2√2+3-2√2=6
x + 1/x =6
Question 2. Let us write which one is greater between (√15+ √3) and (√10+ √8).
Solution: Now, (√15+√3)2 =(√15)2+(√3)2+2.√15.√3
= 15+ 3 + 2√45
= 18+ 2√45
(√10 + √8)2 = (√10)2+(√8)2 +2. √10 √8
= 10 +8 +2√80
= 18+2√80
As 2√80 is greater than 2√45.
(√10+√8)2 > (√15+√3)2
(√10+√8) is greater than (√15+√3).
“WBBSE Class 10 Chapter 9 Quadratic Surd Exercise 9.3 problem-solving steps”
Question 3. Let us write two quadratic surds whose product is a rational number.
Solution: (5+2√6) & (5-2√6).
Question 4. Let us write what should be subtracted from √72 to get √32.
Solution. Required number = √72-√32 = √6x6x2 – √4x4x2
=6√2 -4√2
= 2√2.
Question 5. Let us write the simplified value of (1/√2+1 + 1/√3+ √2 + 1/√4+ √3)
Solution: 1(√2-1)/(√2+1) (√2-1) + 1(√3-√2)/(√3 + √2) (√3-√2)+1(√4-√3)/(√4+ √3)(√4-√3)
=√2-1/2 -1 + √3-√2/3-2 + √4-√34-3
=√2-1+√3-√2+√4-√3
= √4-1
=2-1
= 1.
(1/√2+1 + 1/√3+ √2 + 1/√4+ √3) = 1