Question 1:
Find the nature of the roots of the following quadratic equations.
If the real roots exist, find them;
(I) 2x2 −3x + 5 = 0
(II)
(III) 2x2 − 6x + 3 = 0
If the real roots exist, find them;
(I) 2x2 −3x + 5 = 0
(II)
We know that for a quadratic equation ax2 + bx + c = 0, discriminant is b2 − 4ac.
(A) If b2 − 4ac > 0 → two distinct real roots
(B) If b2 − 4ac = 0 → two equal real roots
(C) If b2 − 4ac < 0 → no real roots
(I) 2x2 −3x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = −3, c = 5
Discriminant = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 − 40
= −31
As b2 − 4ac < 0,
Therefore, no real root is possible for the given equation.
(II)
Comparing this equation with ax2 + bx + c = 0, we obtain
Discriminant
= 48 − 48 = 0
As b2 − 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be
.
and
Therefore, the roots are
.
and
(III) 2x2 − 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = −6, c = 3
Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3)
= 36 − 24 = 12
As b2 − 4ac > 0,
Therefore, distinct real roots exist for this equation as follows.
Therefore, the roots are
.
or
(A) If b2 − 4ac > 0 → two distinct real roots
(B) If b2 − 4ac = 0 → two equal real roots
(C) If b2 − 4ac < 0 → no real roots
(I) 2x2 −3x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = −3, c = 5
Discriminant = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 − 40
= −31
As b2 − 4ac < 0,
Therefore, no real root is possible for the given equation.
(II)
Discriminant
As b2 − 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be
Therefore, the roots are
(III) 2x2 − 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = −6, c = 3
Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3)
= 36 − 24 = 12
As b2 − 4ac > 0,
Therefore, distinct real roots exist for this equation as follows.
Therefore, the roots are
Question 2:
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(I) 2x2 + kx + 3 = 0
(II) kx (x − 2) + 6 = 0
(I) 2x2 + kx + 3 = 0
(II) kx (x − 2) + 6 = 0
We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant
(b2 − 4ac) will be 0.
(I) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we obtain
a = 2, b = k, c = 3
Discriminant = b2 − 4ac = (k)2− 4(2) (3)
= k2 − 24
For equal roots,
Discriminant = 0
k2 − 24 = 0
k2 = 24
(II) kx (x − 2) + 6 = 0
or kx2 − 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = k, b = −2k, c = 6
Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6)
= 4k2 − 24k
For equal roots,
b2 − 4ac = 0
4k2 − 24k = 0
4k (k − 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’.
Therefore, if this equation has two equal roots, k should be 6 only.
(b2 − 4ac) will be 0.
(I) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we obtain
a = 2, b = k, c = 3
Discriminant = b2 − 4ac = (k)2− 4(2) (3)
= k2 − 24
For equal roots,
Discriminant = 0
k2 − 24 = 0
k2 = 24
(II) kx (x − 2) + 6 = 0
or kx2 − 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = k, b = −2k, c = 6
Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6)
= 4k2 − 24k
For equal roots,
b2 − 4ac = 0
4k2 − 24k = 0
4k (k − 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’.
Therefore, if this equation has two equal roots, k should be 6 only.
Question 3:
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2?
If so, find its length and breadth.
If so, find its length and breadth.
Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)
= 2l2
Comparing this equation with al2 + bl + c = 0, we obtain
a = 1 b = 0, c = 400
Discriminant = b2 − 4ac = (0)2 − 4 × (1) × (− 400) = 1600
Here, b2 − 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)
= 2l2
Comparing this equation with al2 + bl + c = 0, we obtain
a = 1 b = 0, c = 400
Discriminant = b2 − 4ac = (0)2 − 4 × (1) × (− 400) = 1600
Here, b2 − 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
Question 4:
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Let the age of one friend be x years.
Age of the other friend will be (20 − x) years.
4 years ago, age of 1st friend = (x − 4) years
And, age of 2nd friend = (20 − x − 4)
= (16 − x) years
Given that,
(x − 4) (16 − x) = 48
16x − 64 − x2 + 4x = 48
− x2 + 20x − 112 = 0
x2 − 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 1, b = −20, c = 112
Discriminant = b2 − 4ac = (− 20)2 − 4 (1) (112)
= 400 − 448 = −48
As b2 − 4ac < 0,
Therefore, no real root is possible for this equation and hence, this situation is not possible.
Age of the other friend will be (20 − x) years.
4 years ago, age of 1st friend = (x − 4) years
And, age of 2nd friend = (20 − x − 4)
= (16 − x) years
Given that,
(x − 4) (16 − x) = 48
16x − 64 − x2 + 4x = 48
− x2 + 20x − 112 = 0
x2 − 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 1, b = −20, c = 112
Discriminant = b2 − 4ac = (− 20)2 − 4 (1) (112)
= 400 − 448 = −48
As b2 − 4ac < 0,
Therefore, no real root is possible for this equation and hence, this situation is not possible.
Question 5:
Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.
Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 − l
Area = l × b = l (40 − l) = 40l − l2
40l − l2 = 400
l2 − 40l + 400 = 0
Comparing this equation with
al2 + bl + c = 0, we obtain
a = 1, b = −40, c = 400
Discriminate = b2 − 4ac = (− 40)2 −4 (1) (400)
= 1600 − 1600 = 0
As b2 − 4ac = 0,
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,
Therefore, length of park, l = 20 m
And breadth of park, b = 40 − l = 40 − 20 = 20
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 − l
Area = l × b = l (40 − l) = 40l − l2
40l − l2 = 400
l2 − 40l + 400 = 0
Comparing this equation with
al2 + bl + c = 0, we obtain
a = 1, b = −40, c = 400
Discriminate = b2 − 4ac = (− 40)2 −4 (1) (400)
= 1600 − 1600 = 0
As b2 − 4ac = 0,
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,
Therefore, length of park, l = 20 m
And breadth of park, b = 40 − l = 40 − 20 = 20