# TEST ON QUADRATIC EQUATIONS FOR CA FOUNDATION

Problem 1 :

Solving equation x2 - (a + b)x + ab = 0, we find the value(s) of x ?

(a)  a      (b)  b   (c) a, b    (d)  None of these

Solution

Problem 2 :

Solving the equation z + √z = 6/25, then the value of z.

(a)  1/5      (b)  2/5     (c) 1/25      (d)  2/25

Solution

Problem 3 :

If

(x + 2)/(x - 2) - (x - 2)/(x + 2) = (x - 1)/(x + 3) - (x + 3)/(x - 3)

then the value of x are

(a)  0, ±√6      (b)   0, ±√3     (c)  0, ±2√3       (d)  None

Solution

Problem 4 :

Solve (x - 1/x)2 + 2(x + 1/x) = 7 1/4

Solution

Problem 5 :

The solution of the equation x - √(25 - x2) = 1

(a)  x = -3     (b)  x =  ±5     (c)  x = 1      (d)  x = 4

Solution

Problem 6 :

Solve the equation

(6x + 2)/4 + (2x2 - 1)/(2x2 + 2) = (10x - 1)/4x we get roots as

(a)  x = -1     (b)  x =  ±1     (c)  x = 1      (d)  x = 0

Solution

Problem 7 :

If α and β  are the roots of

x2 = x + 1

then the value of α2/β - β2/α is

(a)  2√5      (b)  √5     (c)  3√5      (d)  -2√5

Solution

Problem 8 :

Solving x2 + y2 - 25 = 0 and x - y - 1 = 0, we get the roots

(a)  ±3, ±4     (b)  ±2, ±3     (c)  0, 3,  4      (d)  0, -3, -4

Solution

Problem 9 :

The roots of a ax2 + bx + c = 0, are real and unequal if

(a) b2< 4ac     (b) b2 = 4ac   (c) b> 4ac    (d) None

Solution

Problem 10 :

α, β are the roots of the equation x2 - 5x + 6 = 0 the equation with the roots (α β + α + β) and (α β - α - β) is

(a) x- 12x + 11 = 0    (b) 2x- 6x + 12 = 0

(c) x- 12x - 12 = 0    (d) None

Solution

Problem 11 :

If p ≠ q and p2 = 5p - 3 and q2 = 5q - 3 the equation having roots as p/q and q/p is

(a) x- 19x + 3 = 0    (b) 3x- 19x - 3 = 0

(c) 3x- 19x + 3 = 0    (d) 3x+ 19x + 3 = 0

Solution

Problem 12 :

The rational root of the equation 2x3 - x2 - 4x + 2 = 0 is

(a) 1/2    (b) -1/2    (c)  2    (d)  -2

Solution

Problem 13 :

The satisfying value of x3 + x2 - 20x = 0 are

(a) (1, 4, -5)    (b) (2, -4,-5)   (c)  (0, -4, 5)    (d)  (0, 4, -5)

Solution

Problem 14

Solve x3 - 6x2 + 5x + 12 = 0 given that the product of two roots is 12

(a) (1, 3, 4)    (b) (-1,3 ,4)   (c)  (1,6,2 )    (d)  (1, -6, -2)

Solution

Problem 15 :

When √(2z + 1) + √(3z + 4) = 7 the value of z is given by

(a) 1    (b) 2   (c)  3    (d)  4

Solution

1) x = a and x = b, option c

2) t = 3/5 and t = 2/5, option b

3) x = 0 and x = ±√6, option a

4) x = 2 and x = 1/2, x = (-9 ± √65) / 4, option a

5)  x = 4 and x = -3, option a and d

6)  x = 1, option c

7)  -2√5, option d

8)  x = -3, x = 4, option a

9)  b> 4ac, option c

10)  x2 - 12x + 11 = 0, option a

11)  3x2 - 19x + 3 = 0, option c

12)  x = √2, -√2 and 1/2. Option a.

13)  (-5, 0, 4), option d.

14)  3, -1 and 4, option b

15)  z = 4, option d

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