Page 31 - mathsvol1ch1to3ans
P. 31
31
1 1 1 1 1
9. Simplify √ − √ √ + √ √ − √ √ + √ .
3 − 8 8 − 7 7 − 6 6 − 5 5 − 2
Solution: Note that the given expression is
1 1 1 1 1
√ √ − √ √ + √ √ − √ √ + √ √ . Multiplying both numerator and
9 − 8 8 − 7 7 − 6 6 − 5 5 − 4 √ √
denominator by the conjugate of the denominator for each fraction, we have, (3 + 8) − ( 8 +
√
√
√
√
√
√
7) + ( 7 + 6) − ( 6 + 5) + ( 5 + 2) = 3 + 2 = 5
√ √ x + 1
2
10. If x = 2 + 3 find .
2
x − 2
Solution: √ √ √ √
2
2
x + 1 = ( 2 + 3) + 1 = 5 + 2 6 + 1 = 6 + 2 6
Not For Sale - Veeraragavan C S veeraa1729@gmail.com
√ √ √ √
2
2
x − 2 = ( 2 + 3) − 2 = 5 + 2 6 − 2 = 3 + 2 6
√ √ √ √
6 − 2 6 (6 − 2 6)(3 − 2 6) 18 − 6 6 − 24
√ = √ √ =
3 + 2 6 (3 + 2 6)(3 − 2 6) −24
√
−6 − 6 6
=
−15
√
6(1 + 6)
= −
15
√
2(1 + 6)
= −
√ 5
11. Find the square root of 7 + 2 10.
p √ √
Solution: Let 7 + 2 10 = a + b 10 where a, b are rationals.
√ √
Squaring on both sides, 7 + 2 10 = (a + b 10) 2
√
2
2
= a + 10b + 2ab 10

