Quadratic Equation

How to Determine the Nature of the Roots of a Quadratic Equation?

Example 1: Find the values of k for which the quadratic expression (x – k) (x – 10) + 1 = 0 has integral roots.

Solution:

The given equation can be rewritten as, x² – (10 + k)x + 1 + 10k = 0.

D = b² – 4ac = 100 + k² + 20k – 40k = k² – 20k + 96 = (k – 10)² – 4

The quadratic equation will have integral roots if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.

i.e., (k – 10)² – D = 4

Since the discriminant is a perfect square, the difference between two perfect squares in R.H.S will be 4 only when D = 0 and (k – 10)² = 4.

⇒ k – 10 = ± 2. Therefore, the values of k = 8 and 12.

Example 2: Find the values of k such that the equation p/(x + r) + q/(x – r) = k/2x has two equal roots.

Solution:

The given quadratic equation can be rewritten as:

[2p + 2q – k]x² – 2r[p – q]x + r²k = 0

For equal roots, the discriminant (D) = 0, i.e., b² – 4ac = 0

Here, a = [ 2p + 2q – k ], b = – 2r [ p – q ] and c = r²k

[-2r (p – q)]² – 4[(2p + 2q – k) (r²k)] = 0

r²(p – q)² – r²k(2p + 2q – k) = 0

Since r ≠ 0, (p – q)² – k(2p + 2q – k) = 0

k² – 2(p + q)k + (p – q)²

k = 2(p+q) ± √[4(p + q)² – 4(p – q)]²/2 = -(p + q) ± √4pq

∴ The values of k = (p + q) ± 2√pq = (√p ± √q)²

Example 3: Find the quadratic equation with rational coefficients when one root is 1/(2 + √5).

Solution:

If the coefficients are rational, then the irrational roots occur in conjugate pairs. Therefore, if one root is α = 1/(2 + √5) = √5 – 2, then the other root will be β = 1/(2 – √5) = -√5 – 2.

The sum of the roots α + β = -4 and the product of roots α β =  -1.

Thus, the required equation is x² + 4x – 1 = 0.

Example 4: Form a quadratic equation with real coefficients when one of its roots is (3 – 2i).

Solution:

Since the complex roots always occur in pairs, the other root is 3 + 2i. Therefore, by obtaining the sum and the product of the roots, we can form the required quadratic equation.

The sum of the roots is

(3 + 2i) + (3 – 2i) = 6. The product of the root is (3 + 2i) × (3 – 2i) = 9 – 4i² = 9 + 4 = 13.

Hence, the equation is x² – Sx + P = 0

Therefore, x² – 6x + 13 = 0 is the required quadratic equation.

Relationship between Coefficient and Roots of  Quadratic Equation

If α and β are roots of a Quadratic Equation ax² + bx + c = 0, then,

• α + β = -b/a
• αβ = c/a
• α – β = ±√[(α + β)² – 4αβ]
• |α + β| = √D/|a|

The relationship between the roots and coefficient of a polynomial equation can be derived by simplifying the given polynomials and substituting the above results, as shown below.

• α²β + β²α = αβ (α + β) = – bc/a²
• α² + αβ + β² = (α + β)² – αβ = (b² – ac)/a²
• α² + β² = (α – β)² – 2αβ
• α² – β² = (α + β) (α – β)
• α³ + β³ = (α + β)³ – 3αβ(α + β)
• α³ – β³ = (α – β)³ + 3αβ(α – β)
• (α/β)² + (β/α)² = α⁴ + β⁴/α²β²

Example: If the coefficient of x in the quadratic equation x² + bx + c =0 was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original quadratic equation.

Solution:

Since there is no change in the coefficient of x² and c, the product of zeroes will remain the same for both equations.

Therefore, the product of zeroes (c) = -2 × -15 = 30,

Since the original value of b is 13,

Sum of zeroes = -b/a = -13.

Hence, the original quadratic equation is:

x² – (Sum of Zeroes)x + (Product of Zeroes) = 0

x² + 13x + 30 = 0

∴ (x + 10) (x + 3) = 0

Therefore, the roots of the original quadratic equations are -3 and -10.

Quadratic Equations Having Common Roots

Let β be the common root (solution) of quadratic equations a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0. This implies that a₁β² + b₁β + c₁ = 0 and a₂β² + b₂β + c₂ = 0.

Now, solving for β² and β we will get:

β²/(b₁c₂ – b₂c₁) = -β/(a₁c₂ – a₂c₁) = 1/(a₁b₂ – a₂b₁) [using determinant method]

Therefore, β² = (b₁c₂ – b₂c₁)/ (a₁b₂ – a₂b₁) . . . . . . . . . . . . . . . . (1)

And, β = (a₂c₁ – a₁c₂)/(a₁b₂ – a₂b₁) . . . . . . . . . . . . . . . . (2)

On squaring equation (2) and equating it with equation (1), we get:

(a₁b₂ – a₂b₁)/(b₁c₂ – b₂c₁) = (a₂c₁ – a₁c₂)²

Hence, it is the required condition for quadratic equations having one common root.

If both the roots of quadratic equations a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 are common then:

a₁/a₂ = b₁/b₂ = c₁/c₂

If α is a repeated root, i.e., the two roots are α, α of equation f(x) = 0, then α will be a root of the derived equation.

f’(x) = 0 where f’(x) = df/dx

If α is a repeated root common in f(x) = 0 and ϕ(x) = 0, then α is a common root both in f’(x) = 0 and ϕ ‘(x) = 0.

How to Solve Quadratic Equations with Common Roots?

Example: For what value of k do both the quadratic equations, 6x² – 17x + 12 = 0 and 3x² – 2x + k = 0, will have a common root?

Solution:

If one of the roots of quadratic equations a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 is common then: (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁) = (a₂c₁ – a₁c₂)² . . . . . . . . . . . . . (1)

Form the given quadratic equations, a₁ = 6, b₁ = -17, c₁ = 12, a₂ = 3, b₂ = -2 and c₂ = k.

On substituting these values in equation (1), we will get:

[(6×-2) – (3×-17)] × [-17k – (-2×12)] = (3×12 – 6k)²

-663k + 936 = 1296 + 36k² – 432k

36k² + 231k + 360 = 0

12k² + 125k + 120 = 0

(4k + 15) (3k + 8) = 0

Therefore, the values of k are -15/4 and -8/3.

Example: Find the values of k such that the quadratic equations x² – 11x + k = 0 and x² – 14x + 2k = 0 have a common factor.

Solution:

Let (x – α) be the common factor of quadratic equations x² – 11x + k = 0 and x² – 14x + 2k  = 0, then x = α will satisfy the given quadratic equations.

Therefore, α² – 11α + k = 0 . . . . . . . . . . (1)

And, α² – 14α + 2k = 0 . . . . . . . . . . . . (2)

On solving Equation (1) and Equation (2), we will get:

α²/(-22k + 14k) = -α/2k – k = 1/(-14 + 11)

Therefore, α² = (-22k + 14k)/-3 = 8k/3 . . . . . . . (3)

And, α = (2k – k)/(-14 + 11) = k/3 . . . . . . . . . . . (4)

On equating Equation (3) and Equation (4):

8k/3 = (k/3)²

Therefore, the value of k = 24.

Solving Quadratic Equations

There are two methods to solve a quadratic equation

• Algebraic Method
• Graphical Method

Algebraic Method of Solving Quadratic Equations

General Form: ax² + bx + c = 0;

x² + bx/a + c/a = 0

⇒ (x + b/2a)² = b²/4a² – c/a

Or, (x + b/2a)² = (b² – 4ac)/4a²

Or, x + b/2a = ± (√b² – 4ac)/2a

⇒ x = [-b ± √(b² – 4ac)]/2a

b²  – 4ac = Discriminant (D)

• α = (-b+√D)/2a
• β = (-b – √D)/2a

α+β= -b/a, α.β = c/a

Therefore, the quadratic equation can be written as,

⇒ x² – (α + β)x + (α.β) = 0.

Tips to Solve Equations Reducible to Quadratic

• To solve the equations of type ax⁴ + bx² + c = 0, put x² = y
• To solve a.p(x)² + b.p(x) + c = 0, put p(x) = y.
• To solve a.p(x) + b/p(x) + c = 0, put p(x) = y.
• To solve a(x² + 1/x²) + b(x + 1/x) + c = 0,put x + 1/x = y and to solve a(x² + 1/x²) + b(x – 1/x) + c = 0, put x – 1/x = y.
• To solve a reciprocal equation of the type ax⁴ + bx³ + cx² + bx + a = 0, a ≠ 0, divide the equation by d²y/dx² to obtain a(x² + 1/x²) + b(x + 1/x) + c = 0,and then put x + 1/x = y.
• To solve (x + a) (x + b) (x + c) (x + d) + k = 0 where a + b = c + d, put x² +(a + b)x = y
• To solve an equation of type √(ax + b) = cx + d or √(ax² + bx + c) = dx + e, square both sides.
• To solve √(ax + b) ± √(cx + d) = e, transfer one of the radicals to the other side and square both sides. Keep the expression with a radical sign on one side and transfer the remaining expression to the other side.

Graphical Solution of Quadratic Equation

Consider a quadratic equation  ax² + bx + c = 0, where a, b, and c are real and a ≠ 0. The expression can be further rewritten as:

a[(x + b/2a)² – (D/4a²)] = 0

The above quadratic equation represents a parabola whose vertex is at P [-b/2a, -D/4a] and whose axis is parallel to the y-axis.

In a quadratic equation, the value of ‘a’ determines whether the graph of a quadratic equation will be concave upwards (a > 0) or concave downwards (a < 0). The value of discriminant (b² – 4ac) determines whether the graph of a quadratic equation will:

• Intersect the x-axis at two points, i.e. b² – 4ac > 0
• Just touches the x-axis, i.e. b² – 4ac = 0
• Never intersects the x-axis, i.e. b² – 4ac < 0

Graphing Quadratic Equations

Case 1: When a > 0 and b² – 4ac > 0
The graph of a quadratic equation will be concave upwards and will intersect the x-axis at two points α and β with α < β. The quadratic equation will have two real roots (α and β), and the curve will always lie above the x-axis.

• The quadratic function f(x) will be positive, i.e., f(x) > 0, for the values of x lying in the interval (-∞, α) ∪ (β, ∞)
• The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
• The quadratic function f(x) will be negative, i.e., f(x) < 0, for the values of x lying in the interval (α, β)

Case 2: When a > 0 and  b² – 4ac = 0

The graph of a quadratic equation will be concave upwards and will touch the x-axis at a point -b/2a. The quadratic equation will have two equal real roots, i.e., α = β. The quadratic function f(x) will be positive, i.e., 0 ≤ f(x), x ∈ R.

Case 3: When a > 0 and b² – 4ac < 0
The graph of a quadratic equation will be concave upwards and will not intersect the x-axis. The quadratic equation will have imaginary roots, and the curve will always lie above the x-axis. The quadratic function f(x) will be positive, i.e., f(x) > 0, x ∈ R.

Case 4: When a < 0 and  b² – 4ac > 0

The graph of a quadratic equation will be concave downwards and will intersect the x-axis at two points α and β with α < β. The quadratic equation will have two real roots (α and β), and the curve will always lie below the x-axis.

• The quadratic function f(x) will be positive, i.e., f(x) > 0, for the values of x lying in the interval (α, β).
• The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
• The quadratic function f(x) will be negative, i.e., f(x) < 0 for the values of x lying in the interval (−∞, α) ∪ (β, ∞)

Case 5: When a < 0 and b² – 4ac = 0

The graph of a quadratic equation will be concave downwards and will touch the x-axis at a point -b/2a. The quadratic equation will have equal real roots, i.e. α = β. The quadratic function f(x) will be negative, i.e. f(x) ≤ 0, x ∈ R.

Case 6: When a < 0 and b² – 4ac < 0

The graph of a quadratic equation will be concave downwards and will not intersect the x-axis. The quadratic equation will have imaginary roots, and the curve will always lie below the x-axis. The quadratic function f(x) will be negative, i.e. f(x) < 0, x ∈ R.

Range of Quadratic Equation

Consider a quadratic expression f(x) = ax² + bx + c, where a ≠ 0 and a, b, and c are real. The quadratic expression can be further rewritten as f(x) = x² + bax = ca.

 

Case 1: Both the roots of a quadratic equation (α, β) are greater than any given number ‘m’ if,

• b² – 4ac = (D) ≥ 0,
• -b/2a > m,
• f (m) > 0.

Case 2: Both the roots of a quadratic expression (α, β) are less than any given number ‘m’ if,

• b² – 4ac = (D) ≥ 0,
• -b/2a < m,
• f(m) > 0.

Case 3: Both the roots of a quadratic expression (α, β) will lie in the given interval (m1, m2) if,

• b² – 4ac = (D) ≥ 0,
• m₁ < -b/2a > m₂,
• f(m₁) > 0,
• f(m₂) > 0.

Case 4: If exactly one root of a quadratic equation (α, β) will lie in the given interval (m₁, m₂) if, f(m₁).f(m₂) < 0
Case 5: The given number ‘m’ will lie between the roots of a quadratic equation α and β if f(m) < 0

Case 6: The roots of a quadratic equation α and β will have an opposite sign if f(0) < 0.

Case 7: Both the roots of a quadratic expression α and β are positive if,

• b² – 4ac = (D) ≥ 0,
• α + β = -b/a > 0,
• αβ = c/a > 0.

Case 8: Both the roots of a quadratic expression α and β are negative if,

• b² – 4ac = (D) ≥ 0,
• 2. α + β = -b/a < 0,
• 3. αβ = c/a < 0.

How to Find the Range of a Quadratic Equation?

Example 1: Find the range of k for which 6 lies between the roots of the quadratic equation x² + 2(k – 3)x + 9 = 0.

Solution:

6 will lie between the roots of the quadratic expression f(x) = x² + 2(k – 3)x + 9 if,

f (6) < 0

i.e. 36 + 2 (k – 3) . 6 + 9 < 0,

36 + 12k – 36 + 9 < 0,

k < -3/4

Therefore, the range of k for which 6 lies between the roots of the given quadratic equation is:

k ∈ (-∞, -3/4)

Example 2: Find the values of k for which exactly one root of the quadratic equation 4x² – 4(k – 2)x + k – 2 = 0 lies in (0, 1/2).

Solution:

Exactly one root of the quadratic expression 4x² – 4(k – 2)x + k – 2 = 0 will lie in the given interval if,

f(m₁) . f(m₂) < 0

i.e., f(0) . f(1/2)<0

Or, (k – 2) [1 – 2 (k – 2) + k – 2] < 0,

Or, (k -2) (3 – k) < 0,

Or, (k – 2) (k – 3) > 0.

Therefore, the values of k for which exactly one root of the given quadratic equation will lie in the interval (0, 1/2):

k ∈ (-∞, 2) ∪ (3, ∞)

Example 3. Find the values of k for which the roots of the quadratic equation x² – (k – 3)x + k = 0 are greater than 2.

Solution:

The roots of the given quadratic expression f(x) = x² – (k – 3)x + k are greater than 2 if,

Condition 1: -b/2a > 2

i.e., +(k – 3)/2 >2

Therefore, k > 7 . . . . . . . . . . . . . (1)

Condition 2: f (2) > 0

i.e., 4 – (k – 3)² + k > 0,

Or, 4 – 2k + 6 + k > 0.

Therefore, k < 10 . . . . . . . . . . . . . . (2)

Condition 3: b² – 4ac ≥ 0

i.e., (k – 3)² – 4k ≥ 0,

Or, k² + 10k + 9 ≥ 0,

Or, (k – 9) (k – 1) ≥ 0.

Therefore, k ∈ (-∞, 1] ∪ [9, ∞) . . . . . . . . . (3)

From equations (1), (2) and (3), the range of ‘k’ for which the roots of the given quadratic expression are greater than 2: k ∈ [9, 10).

Maximum and Minimum Value of Quadratic Equation

To find the minimum value of a quadratic equation, we need to understand the nature of the graph of these equations for different values of ‘a’. The graph of the quadratic equation f(x) = ax² + bx + c will be either concave upwards (a>0) or concave downwards (a<0), respectively.

When the graph is concave upwards, its vertex determines the minimum value of a quadratic function f(x), and when it is concave downwards, its vertex determines the maximum values of a quadratic function f(x).

How to Find Maximum and Minimum Values of Quadratic Functions?

Case 1: When a > 0, the absolute range of a quadratic equation is given by:

[-b² – 4ac/4a, ∞) or [-D/4a, ∞)

Case 2: When a < 0, the absolute range of a quadratic equation is given by:

(-∞, -D/4a]

Also, the maximum and minimum values of a quadratic equation f(x) occur at x = -b/2a.

If the given quadratic equation is in form f(x) = a(x – h)² + m, then the value of ‘m’ (vertex) gives us the minimum (when ‘a’ is negative) or maximum (when ‘a’ is positive) values of the given function.

Maximum and Minimum Values in Restricted Domain

Case 1: -b/2a ∉ [m, n]

f(x) = ∈ [ minimum {f(m), f(n)}, maximum {f(m), f(n)} ]

Case 2: -b/2a ∈ [m, n]

f(x) ∈ [minimum {f(m), f(n) – D/4a}, maximum {f(m), f(n), -D/4a}]

Example 1: Find the maximum or minimum value of quadratic expression -4(x – 2)² + 2.

Solution:

Since the value of ‘a’ is negative, the given quadratic equation will have a maximum value. Hence, the maximum value of the quadratic equation -4(x – 2)² + 2 is 2.

Example 2: Find the minimum and maximum values of quadratic expression f(x) = x² – 12x + 11.

Solution:

Since a > 0, the maximum and minimum values of a quadratic expression are given by:

[(-b² – 4ac)/4a, ∞) or [-D/4a, ∞)

Therefore, the minimum value of f(x) is:

-(144 – 44)/4 at x = -(-12/2) = -25 at x = 6.

The maximum value of f(x) is infinity.

Therefore, the range of the given quadratic equation is [- 25, ∞).

Example 3: Find the minimum value of equation y = (2x – 5)/(2x² + 3x + 6).

Solution:

f(x) will have minimum value at x = -b/2a [Since a>0]

i.e. at x = -3/4.

∴ The minimum value of quadratic equation f(x) = -D/4a = -(9 – 48)/8 = 39/8.

Example 4: Find the range of function f(x) = (x + 2)/(2x² + 3x + 6), if x is real.

Solution:

Let y = (x + 2)/ (2x² + 3x + 6)

2x²y + 3xy + 6y = x + 2

2x²y + (3y – 1)x + 6y -2 = x + 2

Now, discriminant = (3y – 1)² – 4(2y).(6y – 2) ≥ 0 [because x is real]

9y² + 1 – 6y – 48y² + 16y ≥ 0

39y² – 10y – 1 ≤ 0

39y² – 13y + 3y – 1 ≤ 0

13y(3y – 1) + 1(3y – 1) ≤ 0

∴ The range of quadratic function f(x) = (x+2)/(2x² + 3x + 6): y ∈ [-1/13, 1/3].

⇒ Try this:

If both the roots of the quadratic equation x² + x(4 – 2k) + k² – 3k – 1 = 0 are less than 3, then find the range of values of k.

Ans: k ∈ (-∞, 4).

Descartes’s Rule of Signs for Quadratic Polynomials

This is a very famous rule that helps in getting an idea about the roots of a polynomial equation. The rule states that the number of positive real roots of P_n(x) = 0 cannot be more than the number of sign changes. Similarly, the number of negative roots cannot be more than the number of sign changes in P_n(-x).

Example 1: How many real roots does the equation 2x⁵ + 2x⁴ – 11x³ + 9x² – 4x + 2 = 0 will have?

Solution:

The given equation has 4 sign changes, so it can have a maximum of 4 positive real roots. Now, for f(-x), the equation has only one sign change, i.e. f (-x) = -x⁵ + 2x⁴ + x³ + x² + x + 2 = 0. Hence, the equation will have only one negative real root.

Example 2: How many real roots does the quadratic equation x² + 3|x| + 2 will have?

Solution:

Since the quadratic equation has no sign change for both f(x) and f(-x), the equation will have no real root.

Sign Convention of Quadratic Equation – ax² + bx + c = 0

• The roots of a quadratic equation are equal in magnitude but opposite in sign if b = 0 and ac < 0
• The root with greater magnitude is negative if the sign of a = sign of b × sign of c
• If a > 0, c < 0 or a > 0, c > 0; the roots of quadratic equation will have opposite sign
• If y = ax² + bx + c is positive for all real values of x, a > 0 and D < 0
• If y = ax² + bx + c is negative for all real values of x, a < 0 and D < 0

Example: For what values of ‘m’ does the quadratic expression x² + 2(m + 1)x + 9m – 5 = 0 have only negative roots?

Solution:

Here a = 1, b = 2(m + 1), and c = 9m – 5

Therefore, D = 4(m + 1)2 – 4(9m – 5) = 4m² + 24 – 28m . . . . . (1)

α + β = -2 (m + 1) and αβ = 9m – 5 . . . . . . . . . . . . . . (2)

Now, both the roots of a quadratic expression are negative if b² – 4ac = (D) ≥ 0, α + β = −ba < 0, and αβ = ca < 0

Therefore, 4m² + 24 – 28m ≥ 0 [From equation (1)]

i.e., m² – 7m + 6 ≥ 0

Or, (m – 1) (m – 6) ≥ 0

Therefore, m ≤ 1 or m ≥ 6 . . . . . . . . . . (3)

Now, α + β < 0

i.e., -2(m + 1) < 0 [From equation (2)]

Or, m > -1 . . . . . . . (4)

And, α β < 0

i.e., 9m – 5 < 0

Or, m < 5/9 . . . . . . (5)

From equations (3), (4) and (5), we can conclude that the given quadratic equation will have only negative roots if m ≥ 6.

Quadratic Equations in Two Variables

The general form of a quadratic equation ax² + 2hxy + by² + 2gx + 2fy + c can be factorised into two linear factors as shown below,

ax² + 2(hx + g)x + by² + 2fy + c = 0  . . . .    (i)

Using the quadratic formula in the above equation, we get,

x = [2(hy + g) ± √{4(hy + g)² – 4a(by² + 2fy + c)}]/2a

ax + hy + g = ± √(h²y²  + g² + 2ghy – aby²  – 2afy – ac) . . . . (ii)

At this point, the expression (i) can be resolved into two linear factors if,

(h² – ab)y² + 2(gh – af)y + g² – ac is a perfect square and h² – ab > 0

But (h² – ab)y² + 2(gh – af)y + g² – ac will be a perfect square if discriminant = 0

i.e. g²h² + a²f² – 2afgh – h²g² + abg² + ach² – a²bc = 0 and h² – ab > 0

abc + 2fgh – af² – bg² – ch² = 0 and h² – ab > 0

This is the required condition. The condition that this expression may be resolved into two linear rational factors is:

abc + 2fgh – af² – bg² – ch² = 0 and h² – ab > 0

This expression is called the discriminant of the above quadratic expression.

Biquadratic Equation

A polynomial equation of degree four without the terms of degree three and one is known as a biquadratic equation.

General Form: z⁴ + a₀z² + c = 0

Examples:

• x⁴ +16x² + 6 = 0
• 2x⁴ + 14x² + 3 = 0

How to Solve Biquadratic Equations?

Biquadratic equations can be easily solved by converting them into quadratic equations, i.e. by replacing the variable ‘z’ with x².

Example: Find the zeroes of a biquadratic equation x⁴ – 3x² + 2 = 0.

Solution:

Given x⁴ – 3x² + 2 = 0

On substituting x² = z in the given equation, we get,

z² – 3z + 2 = 0

z² – 2z – z + 2 = 0

z(z – 2) -1(z – 2) = 0

∴ z = 1 and z = 2

Hence, x = ±√1 and x = ±√2 [Since, z = x²].

Practice Questions for Solving Quadratic Equations

Question 1: If G(x) = px² + qx + r and H(x) = -px² + mx + r and p.r ≠ 0, prove that G(x).H(x) = 0 will have exactly two real roots.

Solution:

G(x) will have real roots if q² – 4pr ≥ 0, and H(x) will have real roots if m² + 4pr ≥ 0.

Case 1: If pr < 0,

q² – 4pr ≥ 0, therefore, G(x) will have real root(s).

Case 2: If pr > 0,

m² + 4pr ≥ 0, therefore, H(x) will also have real root(s).

Hence, at least two roots of G(x).H(x) are real.

Question 2: Find the range in which the value of the given quadratic expression will lie (x² + 14x + 9)/(x² + 2x + 3), where x is a real number.

Solution:

Let, y = (x² + 14x + 9)/(x² + 2x + 3)

i.e., x² + 14x + 9 = (x²y + 2xy + 3y)

Or, (1 – y)x² + (7 – y)2x + 3(3 – y) = 0

Since x is real, b² – 4ac ≥ 0

i.e. [2(7 – y)]² – 4[1 – y] × 3[3 – y] ≥ 0

Or, -2y² – 2y + 40 ≥ 0

Or, (y + 5) (y – 4) ≤ 0

Therefore, the range in which the value of the given quadratic expression will lie is -5 ≤ y ≤ 4.

Question 3: Find the values of x for which the expression (x² – 4x + 3)/(x² + x + 1) < 0.

Solution:

Let f(x) = x² – 4x + 3 and g(x) = x² + x + 1.

The coefficient of x² in g(x) is positive, and the value of discriminant (D) < 0. Hence, g(x) is positive for all values of x.

Since, f(x)/g(x) < 0, f (x) must be less than 0.

i.e., x² – 4x + 3 < 0

Or, (x – 3) (x – 1) < 0

Therefore, 1 < x < 3.

Question 4: Find the values of ‘m’ for which the roots of the given quadratic equation f(x) = 9x² + (m – 4)x + m/4 satisfy the following given conditions:

1. Both the roots of quadratic polynomial f(x) are real and distinct.
2. Quadratic expression f(x) has equal roots.
3. The roots of the quadratic expression are not real.
4. The roots of the quadratic equation have opposite signs.
5. Roots are equal in magnitude but have opposite signs.
6. Both the roots of the given quadratic equation f(x) are positive.
7. Both of the roots are negative.

Solution:

From the given quadratic equation f(x) = 9x² + (m – 4)/x + m/4,

a = 9, b = (m – 4) and c = m/4

Therefore, D = b² – 4ac = (m – 4)² – 4(9).(m/4)

= m² + 16 – 8m – 9m = m² – 17m + 16

Therefore, D = m² – 17m + 16 = (m – 1) (m – 16)

Solution 1. Both the roots of the quadratic equations are real and distinct if,

D = (b² – 4ac) > 0

i.e., (m – 16) (m – 1) > 0

Note: If ax² + bx + c > 0, D > 0 and a > 0, then x ∈ (-∞, α) ∪ (β, ∞).

Therefore, both the roots of the given quadratic equation are real and distinct if:

m ∈ (-∞, 1) ∪ (16, ∞).

Solution 2. The quadratic expression f(x) has equal roots if, D = 0

i.e., (m – 16) (m – 1) = 0

Therefore, for m = 16 or m = 1, both the roots of the given quadratic equation are equal.

Solution 3. Roots of the quadratic equation are imaginary if D < 0

i.e., (m – 16) (m – 1) < 0

Note: If ax² + bx + c > 0, D < 0 and a > 0, then x ∈ R

Therefore, the roots of the given equation are imaginary if m ∈ (1, 16).

Solution 4. Roots of quadratic expression will have an opposite sign if f (0) < 0

Since f(0) = m

Therefore, the roots of the given equation will have an opposite sign if m ∈ (−∞,0).

Solution 5. The roots of quadratic expression are equal in magnitude but opposite sign if,

Condition 1: Sum of roots = 0

i.e., (m – 4)/9 = 0, Therefore, m = 4 . . . . . . . . . . . . . . . (1)

Condition 2: D = (b² – 4ac) ≥ 0

i.e. (m -1) (m – 16) ≥ 0

Therefore, x ∈ (=∞, 1] ∪ [16, ∞) . . . . . . . . . . . . . . . . (2)

From equation (1) and (2), we conclude that no such value of ‘m’ exist in f(x) when the roots have an opposite sign with equal magnitude.

Solution 6. Both the roots of a given quadratic expression are positive if,

Condition 1: Product of the roots > 0

i.e., m/(4×9) > 0, Therefore, m > 0 . . . . . . . . . . (3)

Condition 2: Sum of the roots > 0

-(m – 4)/9 > 0, Therefore, m < 4 . . . . . . . . . (4)

Condition 3: D = (b² – 4ac) ≥ 0

From equation (2),

x ∈ (-∞, 1] ∪ [16, ∞) . . . . . . . . (5)

Therefore, from equations (3), (4) and (5), both the roots of f(x) are positive if m ∈ (0,1].

Solution 7. Both the roots are negative if,

Condition 1: The Product of roots > 0

i.e., m/(4×9) > 0, m > 0 . . . . . . . . . . (6)

Condition 2: The Sum of roots < 0

-(m – 4)/9 < 0, m > 4 . . . . . . . . . (7)

Condition 3: D = (b² – 4ac) ≥ 0

Using equation (2),

x ∈ (-∞, 1] ∪ [16, ∞) . . . . . . . . (8)

Therefore, from equations (6), (7) and (8), both the roots of quadratic expression f (x) are negative if,

m ∈ [16, ∞].

Question 5: The quadratic equations x² – ax + b = 0 and x² – px + q = 0 have a common root, and the second equation has equal roots. Show that b + q = ap/2.

Solution:

By considering α and β to be the roots of equation (i) and α to be the common root, we can solve the problem by using the sum and product of the roots formula.

The given quadratic equations are

x² – ax + b = 0 ………. (i)

x² – px + q = 0 ………..(ii)

From equation (i), α + β = a, α = b

From equation (ii), 2α = p, α² = q

b + q = αβ + α² = α (α + β) = ap/2

Question 6: x² + ax + bc = 0 and x² + bx + ca = 0 have a non zero common root and a ≠ b. Show that the other roots are roots of the quadratic equation x² + cx + ab = 0, c ≠ 0.

Solution:

By considering α to be the common root of the quadratic equations and β, γ to be the other roots of the equations, respectively, then by using the sum and product of roots formula, we can prove this.

Further, α + β = -a and αβ = bc;

α + γ = -b, αγ = ca

2α + β + γ = -(a + b) and α²βγ = abc² . . . . . . (i)

Therefore, β + γ = c – 2c = -c . . . . . . . (ii)

and, c²βγ = c²ba

Therefore, βγ = ab …. (iii)

From equations (ii) and (iii),

β and γ are the roots of the quadratic equation x² + cx + ab = 0.

Question 7: The expressions x² – 11x + a = 0 and x² – 14x + 2a = 0 must have a common factor and a ≠ 0. Find the common factor and the common root.

Solution:

If (x – α) is the common factor of the given quadratic equations, then x = α becomes the root of the corresponding equation.

Therefore, α² – 11α + a = 0 and α² – 14α + 2a = 0

On subtracting the above equations, we get,

3α – a = 0 ⇒ α = a/3

Hence, a²/9 – 11a/3 + a = 0,

On solving the above quadratic equation, we get, a = 0 or 24

Since a ≠ 0, therefore, a = 24.

Question 8: Determine the values of m for which the quadratic equations 3x² + 4mx + 2 = 0 and 2x² + 3x – 2 = 0 may have a common root.

Solution:

Consider α to be the common root of the given equations.

Then, 3α² + 4mα + 2 = 0 and 2α² + 3α – 2 = 0

Using the cross multiplication method, we have

(-6 – 4)² = (9 – 8m) (-8m – 6)

⇒ 50 = (8m – 9)(4m + 3)

Or, 32m² – 12m – 77 = 0

Or, (8m + 11) (4m – 7) = 0

Therefore, the values of m for which the given quadratic equations will have common roots  = -11/8 and 7/4.

Question 9: Find the relation between c, a and m if a quadratic equation formed from y² = 4ax and y = mx + c has equal roots.

Solution:

(mx + c)² = 4axm²x² + 2(cm – 2a)x + c² = 0

Since the roots of this quadratic equation are equal, D = 04(cm – 2a)² – 4m²c² = 0

4a² – 4cma = 0 or a² = acm

⇒ a = cm or c = a/m.

This is a condition for the line y = mx + c to be tangent to the curve y² = 4ax.

Question 10: For what values of ‘m’ does the quadratic equation (1 + 2) x² – 2(1 + 3m)x + (1 + 8m) = 0 have equal roots?

Solution:

The roots are equal if discriminant (D) = 4(1 + 3m)² – 4(1 + m)(1 + 8m) = 0

m² – 3m = 0 ⇒ m = 0, 3.

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