Questions - Answers
1. If x2+bx+c = 0, x2+cx+b = 0 (b ≠ c) have a common root then show that b+c+1 = 0.
Sol. Let α be the common root. Then
α2 + bα + c = 0 ------------------------- (1)
α2 + cα + b = 0 -------------------------- (2)
(1) - (2) ⇒
(b-c)α + (c-b) = 0 ⇒ α = 1
(1) ⇒ 1 + b + c = 0 or b + c + 1 = 0.
2. Find the nature of the roots of 3x2+7x+2 = 0.
Sol. Here a = 3, b = 7, c = 2.
Δ = b2-4ac = 49-4(3)(2) = 25 = 52
⇒ The roots of the given equation are rational and unequal.
3. Discuss the signs of the Quadratic expressions x2-5x+4, when x is real.
Sol: x2-5x+4 = (x-1) (x-4)
a = 1 > 0, roots are 1, 4.
x2-5x+4 > 0 if x < 1 or x > 4
x2-5x+4 < 0 if 1 < x < 4.
4. If α, β are the roots of the equation ax2+bx+c = 0, find the values of the following expressions in terms of a, b, c.
Sol. Given α, β are roots of ax2+bx+c = 0
5. Prove that the roots of (x-a) (x-b) = h2 are always real.
Sol: Given equation is (x-a)(x-b) = h2
⇒ x2 - (a+b)x + (ab-h2) = 0
Δ = (a+b)2 - 4(1)(ab-h2)
= (a-b)2 + (2h)2 > 0
⇒ The roots are real.
6. Find the changes in the sign of the expression 15+4x-3x2 and find its extreme value.
Sol: 15+4x-3x2 = (3-x) (5+3x).
a = -3 < 0; roots are 3, -5/3
15 + 4x -3x2 > 0 if -5/3 < x < 3
15 + 4x-3x2< 0 if x < -5/3 or x > 3.
The maximum value of 15 + 4x -3x2 is
7. Find the Quadratic equation, the sum of whose roots are 1 and the sum of the squares of the roots are 13.
Sol: Let α, β be the roots. Then α + β = 1, α2 + β2 = 13
αβ = 1/2 [(α + β)2 -(α2 + β2)] = [1 - 13] = -6
The required equation is x2 - x - 6 = 0.
8. For what values of m will the equation (m+1)x2 + 2(m+3)x + (m+8) = 0 has equal roots.
Sol: Since the given equation has equal roots discriminant Δ = 0
ie., b2 - 4ac = 0 Here a = m+1, b = 2(m+3), c = m+8
Δ = 0 ⇒ [2(m+3)]2 - 4 (m+1) (m+8) = 0
4 (m2+9+6m) - 4 (m2+9m+8) = 0
- 12m + 4 = 0
m = 1/3.
9. Find the condition that one root of the Quadratic equation ax2+bx+c = 0 shall be n times the other, where n is a positive integer.
Sol: Let the roots be α, nα.
Sum of roots: α + nα = -b/a
⇒ α (n+1) = -b/a
α = 
Product of roots: (α) (nα) = c/a
nα2 = c/a
nb2 = ac(n+1)2.
1. If the equation x2 − 15 − m(2x − 8) = 0 has equal roots, find the values of m.
Theorem 1: If a quadratic equation ax2 + bx +c = 0 has complex roots ax2 + bx + c and "a" have the same sign when x is a real number.
Proof: Given that ax2 + bx + c = 0 has complex roots, so
ax2 + bx + c and "a" both have the same sign.
Theorem 2: If a and b (a < b) are the real roots of ax2 + bx + c = 0, then
i) ax2 + bx + c and 'a' are of opposite sign when a < x < b
ii) ax2 + bx + c and 'a' are of the same sign when x < a or x > b.
Proof: If a and b are the roots of ax2 + bx + c = 0,
Theorem 3: Suppose that f(x) = ax2 + bx + c is a quadratic expression.
Solved Problems
1. Discuss the signs of the following quadratic expressions when x is real.
2. Find the maximum or minimum of the following expression as x varies over R.
3. Find the changes in the sign of the followingexpressions and find their extreme value.
4. Determine the range of the following expression.
4. Marks Questions
⇒ 3yx2 + (4y-4)x + (y-1) = 0
x ∈ R ⇒ (4y-4)2 - 4(3y) (y-1) ≥ 0
⇒ 16y2 + 16 - 32y - 12y2 + 12y ≥ 0
⇒ 4y2 - 20y + 16 ≥ 0
⇒ y2-5y + 4 ≥ 0
⇒ (y-1) (y-4) ≥ 0
Roots of y2 - 5y + 4 = 0 are 1 & 4
'.' Coefficient of y2 = 1 > 0 & expression ≥ 0
y does not lie between 1 & 4.
⇒ yx2 + (-3y-1)x + (2y+p) = 0
x ∈ IR ⇒ (-3y-1)2 - 4(y) (2y+p) ≥ 0
⇒ 9y2+1 + 6y - 8y2 - 4py ≥ 0
⇒ y2 + (6-4p)y +1 ≥ 0
y is real, y2 + (6-4p) y+1 ≥ 0
The roots are imaginary or real and equal
⇒ (6-4p)2 - 4(1) (1) ≤ 0
⇒ 36 + 16p2 - 48p - 4 ≤ 0
⇒ 16p2 - 48p + 32 ≤ 0
⇒ p2 - 3p + 2 ≤ 0
⇒ (p - 1) (p - 2) ≤ 0
Roots of p2 - 3p + 2 = 0 are 1, 2
∵ Coefficient of p2 = 1 > 0 and expression ≤ 0
1 < p < 2 and
3. If c2 ≠ ab and the roots of (c2-ab)x2 - 2(a2-bc)x + (b2-ac) = 0 are equal, then show that a3+b3+c3 = 3 abc or a = 0.
Sol. Given equation has equal roots
⇒ Discriminant Δ = b2-4ac = 0.
⇒ [-2(a2-bc)]2 - 4 (c2-ab) (b2-ac) = 0
⇒ 4(a4+b2c2-2a2bc) - 4 (c2b2-ac3-ab3+a2bc) = 0
⇒ a4+b2c2-2a2bc-c2b2+ac3+ab3-a2bc = 0
⇒ a4+ab3+ac3-3a2bc = 0
⇒ a (a3+b3+c3-3abc) = 0
⇒ a = 0 or a3+b3+c3-3abc = 0
⇒ a = 0 or a3+b3+c3 = 3 abc.
