Quadratic Equation
Question
Q1.
Solve (x1/3)2 + x1/3 - 2 = 0
(a) (1,-8)
(b) (8,-8)
(c) (1,-1)
(d) (1,8)
Solution. a
Let x1/3 = m
m2 + m - 2 = 0
m2 + 2m - m - 2 = 0
m(m + 2) - 1 (m + 2) = 0
(m - 1)(m + 2) = 0
m = 1, -2
we known that x1/3 = m
So put m = 1
x1/3 = 1
x = 13
Now put m = -2
x1/3 = -2
x = (-2)3 = -8
Solution
Q2.
Find the quadratic equation which has a root 4 - √3
(a) x2 + 8x - 13 = 0
(b) x2 - 8x + 13 = 0
(c) x2 - 8x - 10 = 0
(d) None of these
Solution. b
Let α and β are two roots of the equation , one root is 4 - √3 . So another will be conjugate of this 4 + √3
Required equation is x2 - (α + β)x + α β = 0
(α + β) = 4 - √3 + 4 + √3 = 8
α β = (4 - √3) (4 + √3) = 16 - 3 = 13
x2 - 8x + 13 = 0
Solution
Q3.
The shape of f(x) = ax2 + bx + c = 0 is a __________ .
(a) Circle
(b) Triangle
(c) Parabola
(d) Straight line
Solution.c
Parabola
Note : A polynomial of degree two has parabolic curve .
Solution
Q4.
The number of roots of the equation x3 + x2 - x - 1 = 0 are
(a) 1
(b) 2
(c) 4
(d) 3
Solution. d
Note : Number of roots will be equal to the degree of the equation . This is equation of degree 3.
Solution
Q5.
The number of real solution of y2 - 3lyl + 2 = 0 is
(a) 3
(b) 4
(c) 2
(d) none of these
Solution. b
y2 - 3lyl + 2 = 0
lyl = ±y
lyl = y
∴ y2 - 3lyl + 2 = 0
∴ (y - 2) (y - 1) = 0
∴ y = 2 or y = 1
lyl = -y
∴ y2 + 3lyl + 2 = 0
∴ (y + 2)(y + 1) = 0
∴ y = -2 or y = -1
Solution
