Mathematics
Class-X
2015-2016
Q.1. Solve the equation : 2x – 1/x = 7. Write your answer correct to two decimal places.
Solution :
We have, 2x – 1/x = 7
Or, 2x2 – 1 = 7x
Or, 2x2 – 7x – 1 = 0 ,
Here a = 2, b = – 7, c = – 1, D = b2 – 4ac = (–7)2 – 4×2×–1 = 57.
Using formula , x = [– b ± √D]/2a, we get
x = [– (–7) ± √57] /2×2
= [7 ± √57]/4
= [7 ± 7.550]/4
Therefore, x = [7 + 7.550]/4 or, x = [7 – 7.550]/4
Either, x = 3.64 or, x = – 0.14 [Ans.]
Q.2. Solve the following quadratic equation for x and give your answer correct to two
decimal places : x2 – 3x – 9 = 0
Solution :
We have x2 – 3x – 9 = 0
Comparing with ax2 + bx + c = 0
We get, a = 1, b = – 3, c = – 9 and D = b2 – 4ac = (– 3)2 – 4 (1)(– 9)
= 9 + 36 = 45.
Hence, x = [– b ± √D]/2a = [– (– 3) ± √ (45)] / 2
= [3 ± 3√5]/2 = [3 ± 6.708]/2
Or, x = 4.85 and 0 – 1.85. [Ans.]
Q.3. Solve the following equation and give your answer up to two decimal places :
x2 – 5x – 10 = 0.
Solution :
We have, x2 – 5x – 10 = 0
Here a = 1, b = – 5, c = – 10, D = b2 – 4ac = (– 5)2 – 4×1×(– 10) = 25 + 40 = 65.
As, x = [– b ± √D]/2a = [–(– 5) ± √65]/2
= [5 ± 8.06]/2.
Therefore, x = [5 + 8.06]/2 = 13.02/2 = 6.53 or [5 – 8.06]/2 = – 3.06/2 = – 1.53.
Hence, x = 6.53 ; – 1.53 [Ans.]
Q.4. Solve the equation 3x2 – x – 7 = 0 and give your answer correct to two decimal places.
Solution :
Do yourself [Ans. = 1.703, – 1.37]
Q.5. Solve (7x + 1)/(7x + 5) = (3x + 1)/(5x + 1).
Solution :
Here we have, (7x + 1)/(7x + 5) = (3x +1)/(5x +1)
Multiplying both sides by (7x +5)(5x +1) [LCM of the fraction], we get
(7x +1)(5x +1) = (3x +1)(7x +5)
or, 35x2 + 7x + 5x + 1 = 21x2 + 7x + 15x + 5
or, 35x2 + 12x + 1 = 21x2 + 22x + 5
or, 35x2 – 21x2 +12x – 22x + 1 – 5 = 0
or, 14x2 – 10x – 4 = 0
Sudheer Gupta .
Be positive and constructive. 
Page 1
Mathematics
Class-X
2015-2016
or, 7x2 – 5 x – 2 = 0
or, 7x2 – (7 – 2)x – 2 = 0 [By splitting middle term]
or, 7x2 – 7x + 2x – 2 = 0
or, 7x(x – 1) + 2(x – 1) = 0
or, (x – 1)(7x +2) = 0 [By factorization]
either x – 1 = 0 or, 7x + 2 = 0
either x = 1 or, x = – 2/7
Hence, roots of the quadratic equation are 1, – 2/7. [Ans.]
Q.6. Solve the quadratic equation : 21x2 – 8x – 4 = 0
Solution :
Here we have, 21x2 – 8x – 4 = 0
Or, 21x2 – (14 – 6)x – 4 = 0 [By splitting middle term]
Or, 21x2 – 14x + 6x – 4 = 0
Or, 7x(3x – 2) + 2(3x – 2) = 0
Or, (3x – 2 )(7x + 2) = 0 [By factorization]
Either 3x – 2 = 0 or, 7x + 2 = 0
Either 3x = 2 or, 7x = – 2
Either x = 2/3 or, x = – 2/7
Hence, roots of the quadratic equation are 2/3, – 2/7. [Ans.]
Q.7. Solve the equation : √(3x2 – 2) = 2x – 1.
Solution :
Here we have, √(3x2 – 2) = 2x – 1,
Squaring both sides we get, 3x2 – 2 = (2x – 1)2
Or, 3x2 – 2 = 4x2 – 4x + 1
Or, 3x2 – 4x2 + 4x – 2 – 1 = 0
Or, – x2 + 4x – 3 = 0
Or, x2 – 4x + 3 = 0
Or, x2 – (3 + 1)x + 3 = 0
Or, x2 – 3x – x + 3 = 0
Or, x(x – 3) – (x – 3) = 0
Or, (x – 3)(x – 1) = 0
Either , x – 3 = 0 Or, x – 1 = 0
Either , x = 3 Or, x = 1 . Hence roots are 1,3. [Ans.]
Q.8. Solve the quadratic equation : √3 x2 + 10x – 8√3 = 0.
Solution :
Here we have, √3 x2 + 10x – 8√3 = 0; where, a = √3, b = 10, c = – 8√3.
Using, D = b2 – 4ac = (10)2 – 4 ×√3 ×(– 8√3) = 196 >0
Hence, x = [– b ± √D]/2a = [– 10 ±√(196)]/2×√3
Or, x = [– 10 ±14]/2√3. Or, x = 4/2√3 , 0 – 24 /2√3
Therefore , x = (2√3)/3 , – 4√3. [Ans.]
Q.9. Solve using quadratic formula : x2 – 4x + 1 = 0.
Solution :
Do yourself. [Ans. = 2 – √3, 2 + √3 ]
Sudheer Gupta .
Be positive and constructive. 
Page 2
Mathematics
Class-X
2015-2016
Q.10. Solve for x and give your answer correct to 2 decimal places. x2 – 10x + 6 = 0.
Solution:
Do yourself. [Ans. x = 9.36 and 0.64]
Sudheer Gupta .
Be positive and constructive. 
Page 3