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) b² > 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) x² - 12x + 11 = 0    (b) 2x² - 6x + 12 = 0

(c) x² - 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) x² - 19x + 3 = 0    (b) 3x² - 19x - 3 = 0

(c) 3x² - 19x + 3 = 0    (d) 3x² + 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