Exercise 1(C)
Question 1
(i) √13
(ii) √512
(iii) √14675
Sol :
(i) √13
Let be √13
⇒1×√3√3×√3 (Multiplying and dividing by √3)
=√33
(ii) √512=√5√2×2×3=√52√3
Multiplying and dividing by √3
⇒√5√32√3⋅√3⇒√5√32×3
⇒√5√36
⇒√156=16√15
(iii) √14675
=√1×75+4675=√12175=√121√75=√11×11√3×5×5⇒11√35√3×√3=11√35×3=1115√3
Question 2
√112−√63+224√28
Sol :
⇒√112−√63+224√28
⇒√2×2×2×2×7−√3×3+7+224√2×2×7
⇒2×2√7−3√7+2242√7
⇒4√7−3√7+112√7
⇒4√7−3√7+112√7√7×√7
⇒4√7−3√7+112√77
⇒4√7−3√7+16√7
⇒(4−3+16)√7=17√7
Question 3
4√18√12−8√75√32+9√23
Sol :
4√18√12−8√75√32+9√23
=4√2×3×3√2×2×3−8√3×5×5√2×2×2×2×2+9√2√3
=4×3√22√3−8×5√32×2√2+9√2√3
=6√2√3−10√3√2+9√2√3
=6√2×√3√3×√3−10√3×√2√2×√2+9√2×√3√3×√3
(Rationalising denominator of each)
=6√63−10√62+9√63
=2√6−5√6+3√6
=5√6−5√6=0
Question 4
(i) 14−√3
(ii) 2√5+√3
(iii) 12√5−√3
(iv) √3+√2√3−√2
Sol :
(i) 14−√3×4+√34+√3
=4+√3(4)2−(√3)2
(ii) 2√5+√3=2(√5−√3)(√5+√3)(√5−√3)
Rationalising factor of (√5+√3) is
=2(√5−√3)5−3=2√5+2√32
(iii) 12√5−√3=1(2√5+√3)(2√5−√3)(2√5+√3)
Rationalising factor of (2√5−3) is (2√5+3)
=2√5+√3(2√5)2−(√3)2
(iv) √3+√2√3−√2
=(√3+√2)(√3+√2)(√3−√2)(√3+√2)
Rationalising factor of (√3−√2) is (√3+√2)
=(√3+√2)2(√3)2−(√2)2
=3+2+2√3×√23−2
Question 5
Rationalise the denominator and simplify :
Sol :
(i) 4+√54−√5+4−√54+√5
=(4+√5)(4+55)(4−√5)(4+√5)+(4−√5)(4−√5)(4+√5)(4−√5)
Rationalising the denominator
=(4+√5)2(4)2−(√5)2+(4−√5)2(4)2−(√5)2
=16+5+2×4√516−5+16+5−2×4√516−5
=21+8√511+21−8√511
=21+8√5+21−8√511
=4211
(ii) 35−√3+25+√3
=3(5+√3)(5−√3)(5+√3)+2(5−√3)(5+√3)(5−√3)
Rationalising the denominator
=15+3√325−3+10−2√325−3
=15+3√322+10−2√322
=15+3√3+10−2√322
=25+√322
Question 6
Find the values of a and b if
(i) 3+√23−√2=a=a+b√2
(ii) 5+2√37+4√3=a+b√3
(iii) √7−1√7+1−√7+1√7−1=a+b√7
Sol :
(i) 3+√23−√2=a=a+b√2
=(3+√2)(3+√2)(3−√2)(3+√3)=a+b√2
=(3+√2)29−2=a+b√2
=9+2+6√27=a+b√9
=11+6√27=a+b√2
=117+67√2=a+b√2
comparing
a=117,b=67
(ii) 5+2√37+4√3=a+b√3
=(5+2√3)(7−4√3)(7+4√3)(7−4√3)=a+b√3
(Rationalising the denominator)
=35−20√3+14√3−8×3(7)2−(4√3)2=a+b√3
=35−24−6√349−48=a+b√3
=11−6√31=a+b√3
=11−6√3=a+b√3
we get a=11, b=-6
(iii) √7−1√7+1−√7+1√7−1=a+b√7
=(√7−1)2−(√7+1)2(√7+1)(√7−1)=a+b√7
(7+1−2√7)−(7+1+2√7)7−1=a+b√7
=8−2√7−8−2√26=a+b√7
=−4√76=a+b√7
−23√7=a+b√7
we get
a=0 , b=−23
Question 7
Rationalise the denominator of 1√2+√3+√5
Sol :
1√2+√3+√5=1(√2+√3)+√5
Rationalising denominator
=(√2+√3)−(√5)[(√2+√3)+√5)(√2+√3)−√5]
=√2+√3−√5(√2+√3)2−(√5)2=√2+√3−√52+3+2√6−5
=√2+√3−√52√6
Rationalising
=(√2+√3−√5)√62×√6×√6
=√12+√18−√302×6
=√4×3+√9×2−√3012
=2√3+3√2−√3012
Question 8
Taking √2=1.414 and √3=1.732 , find without using tables or long division, the value of
(a) 13−√2
(b) 2√3−√2
Sol :
(i) √2=1.414,√3=1.732
13−√2=1(3+√2)(3−√2)(3+√2)
Rationalising the denominator
=3+√2(3)2−(52)2
=3+√2(3)2(√2)2
=3+√29−2
=3+√27
=3+1.4147
=4.4147
=0.631
(ii) 2√3−√2=2(√3+√2)(√3−√2)(√3+√2)
Rationalising denominator
=2(√3+√2)(√3)2−(√2)2
=2(√3+√2)3−2
=2(√3+√2)
=2(1.732+1.414)
=2×3.146
=6.292
Question 9
Express 3−5√53+√5 in the form (a√5–b) where a and b are simple fractions.
Sol :
3−5√53+√5=(3−5√5)(3−2√5)(3+2√5)(3−3√5)
=9−6√5−15√5+10×5(3)2−(2√5)2
=9−21√5+209−20
=59−21√5−11
=−5911+2111√5
=It is in the form of a√5−b
Question 10
Prove that 1√2−1+2√3+1=√2+√3
Sol :
L.H.S=1√2−1+2√3+1
=√2+1(√2−1)(√2+1)+2(√3−1)(√3+1)(√3−1)
=√2+1(√2)2−(1)2+2(√3−1)(√3)2−(1)2
=√2+12−1+2(√3−1)3−1
=√2+11+2(√3−1)2
=R.H.S
L.H.S=R.H.S
Question 11
Simplify:
6√2√3+√6−4√3√6+√2+2√6√2+√3
Sol :
=6√2√3+√6−4√3√6+√2+2√6√2+√3
=6√2(√3−√6)(√3+√6)(√3−√6)−4√3(√6−√2)(√6+√2)(√6−√2)+2√6(√2−√3)(√2+√3)(√2−√3)
Rationalising the denominator
=6√2(√3−√6)(√3)2−(√6)2−4√3(√6−√2)(√6)2−(√2)2+2√6(√2−√3)(√2)2−(√3)2
=6√2(√3−√6)3−6−4√3(√6−√2)6−2+2√6(√2−√3)2−3
=6√2(√3−56)−3−4√3(√6−√2)4+2√6(√2−√3)−1
=−2√2(√3−√6)−√3(√6−√2)−2√6(√2−√3)
=−2√6+2√12−√18+√6−2√12+2√18
=−2√6+2×2√3−3√2+√6−2×2√3+2×√3+2×3√2
=−2√6+4√3−3√2+√6−4√3+6√2
=−2√6+√6+4√3−4√3−3√2+6√2
=−√6+3√2
=3√2−√6
=√9×2−√6
=√18−√6
Question 12
Simplify:
(i) 62√3−√6+√6√3+√2−4√3√6−√2
(ii) 7√3√10+√3−2√5√6+√5−3√2√15+3√2
Sol :
(i) 62√3−√6+√6√3+√2−4√3√6−√2
6(2√3+√6)(2√3−√6)(2√3+√6)
=6(2√3+√6)(2√3)2−(√6)2
=6(2√3+√6)12−6
=6(2√3+√6)6
=2√3+√6...(i)
√6√3+√2
=√6(√3−√2)(√3+√2)(√3−√2)
=√6(√3−√2)3−2
=√18−√121
=3√2−2√3...(ii)
4√3√6−√2
=4√3(√6+√2)(√6−√2)(√6+√2)
=4√3(√6+√2)6−2
=4√3(√6+√2)4
=√3(√6+√12)
=√18+√6
=3√2+√6...(iii)
From (i) , (ii) , (iii)
=(2√3+√6)+(3√2−2√3)−(3√2+√6)
=2√3+√6+3√2−2√3−3√2−√6
=0
(ii) 7√3√10+√3−2√5√6+√5−3√2√15+3√2
=7√3√10+√3
=7√3(√10−√3)10−3
=7√3(√10−√3)10−3
=7√3(√10−√3)7
=√3(√10−√3)
=√30−3...(i)
2√5√6+√5=2√5(√6−√5)(√6+√5)(√6−√5)
=2√5(√6−√5)6−5
=2√5(√6−√5)1
=2√30−2×5
=2√30−10...(ii)
=3√2√15+3√2
=3√2(15−3√2)(√15+3√2)(√15−3√2)
=3√2(√15−3√2)15−18
=3√2(√15−3√2)−3
=−√2(√15−3√2)
=−√30+3×2
=−√30+6....(iii)
From (i), (ii) and (iii)
=(√30−3)−(2√30−10)−(−√30+6)
=√30−3−2√30+10+√30−6
=10-9
=1
Question 13
If x=2+√3 , find the value of x2+1x2
Sol :
⇒x=2+√3
⇒1x=12+√3
=2−√3(2+√3)(2−√3)
=2−√34−3
=2−√31
=2−√3
x+1x=2+√3+2−√3
=4
Squaring both side
=(x+1x)2=(4)2
=x2+1x2+2=16
=x2+1x2=16−2
=14
Question 14
If x=√2+1 , find the value of x2+1x2
Sol :
⇒x=√x+1
⇒1x=1√x+1
=√2−1(√2+1)(√2−1)
x+1x=√2+1+√2−1
Squaring both sides
[x+1x]2=(2√2)2
=x2+1x2+2
=4×2=8
=x2+1x=8−2
=6
Question 15
If x=5−√212 , find the value of
(i) x+1x and (ii) x2+1x2
Sol :
(i) x=5−√212
=1x=25−√21
=2(5+√21)(5−√21)(5+√21)
=2(5+√21)25−21
=2(5+√21)4
=5+√212
x+1x=5−√212+5+√212
=5−√21+5+√212
=102
=5
=x+1x=5
Squaring both sides
(x+1x)2=(5)2
=x2+1x2+2
=25
=x2+1x2
=25-2
=23
x+1x=5
x2+1x2=23
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