Inter 2nd Year Maths 2A Quadratic Expressions Formulas

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 3 Quadratic Expressions to solve questions creatively.

Intermediate 2nd Year Maths 2A Quadratic Expressions Formulas

→ If a, b, c are real or complex numbers and a ≠ 0, then the expression ax² + bx + c is called a quadratic expression in the variable x.
Eg: 4x² – 2x + 3

→ If a, b, c are real or complex numbers and a ≠ 0, then ax² + bx + c = 0 is called a quadratic equation in x.
Eg: 2x² – 5x + 6 = 0

→ A complex number α is said to be a root or solution of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0

→ The roots of ax² + bx + c = 0 are [latex]\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/latex]

→ If α, β are roots of ax² + bx + c = 0, then α + β = [latex]\frac{-b}{a}[/latex] and αβ = [latex]\frac{C}{a}[/latex]

→ The equation of whose roots are α, β is x² – (α + β)x + αβ = 0

→ Nature of the roots: ∆ = b² – 4ac is called the discriminant of the quadratic equation ax² + bx + c = 0. Let α, β be the roots of the quadratic equation ax² + bx + c = 0
Case 1: If a, b, c are real numbers, then

∆ = 0 ⇔ α = β = [latex]\frac{-b}{2 a}[/latex] (a repeated root or double root)
∆ > 0 ⇔ α and β are real and distinct.
∆ < 0, ⇔ α and β are non- real complex numbers conjugate to each other.

Case 2: If a, b, c are rational numbers, then

∆ = 0 ⇔ α and β are rational and equal (ei) α = [latex]\frac{-b}{2 a}[/latex], a double root or a repeated root.
∆ > 0 and is a square of a rational number ⇔ α and β are rational and distinct.
∆ > 0 but not a square of a rational number ⇔ α and β are conjugate surds.
∆ < 0, ⇔ α and β are non- real ⇔ α and β are non-real con conjugate complex numbers.

→ Let a, b and c are rational numbers, α and β be the roots of the equations ax² + bx + c = 0. Then

α, β are equal rational numbers if ∆ = 0.
α, β are distinct rational numbers if ∆ is the square of a non zero rational numbers.
α, β are conjugate surds if ∆ > 0 and ∆ is not the square of a nonzero square of a rational number.

→ If a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have two same roots, then [latex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/latex]

→ If α, β are roots of ax² + bx + c = 0,

the equation whose roots are [latex]\frac{1}{\alpha}, \frac{1}{\beta}[/latex] is f [latex]\left(\frac{1}{x}\right)[/latex] = 0. If c ≠ 0 (ie) αβ ≠ 0
the equation whose roots are α + k, β + k is f(x – k) = 0
the equation whose roots are kα and kβ is f[latex]\left(\frac{x}{k}\right)[/latex] = 0
the equation whose roots are equal but opposite in sign is f(-x) = 0
(ie) the equation whose roots are – α, – β is f(-x) = 0.

→ If the roots of ax² + bx + c = 0 are complex roots then for x ∈ R, ax² + bx + c and ‘a’ have the same sign.

→ If α and β (α < β) are the roots of ax² + bx + c = 0 then

ax² + bx + c and ‘a’ are of opposite sign when α < x < β
ax² + bx + c and ‘a’ are of the same sign if x < α or x > β.

→ Let f(x) = ax² + bx + c be a quadratic function

If a > 0 then f(x) has minimum value at x = [latex]\frac{-b}{2 a}[/latex] and the minimum value is given by [latex]\frac{4 a c-b^{2}}{4 a}[/latex]
If a < 0 then f(x) has maximum value at x = [latex]\frac{-b}{2 a}[/latex] and the maximum value is given by [latex]\frac{4 a c-b^{2}}{4 a}[/latex]

→ A necessary and sufficient condition for the quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).

→ If a₁b₂ – a₂b₁ = 0 then common root of a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 is [latex]\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}[/latex].

→ The standard form of a quadratic ax² + bx + c = 0 where a, b, c ∈ R and a ≠ 0

→ The roots of ax² + bx + c = 0 are [latex]\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/latex]

→ For the equation ax² + bx + c = 0, sum of the roots = [latex]-\frac{b}{a}[/latex], product of the roots = [latex]\frac{c}{a}[/latex].

→ If the roots of a quadratic are known, the equation is x² – (sum of the roots)x +(product of the roots)= 0

→ “Irrational roots” of a quadratic equation with “rational coefficients” occur in conjugate pairs. If p + √q is a root of ax² + bx + c = 0, then p – √q is also a root of the equation.

→ “Imaginary” or “Complex Roots” of a quadratic equation with “real coefficients” occur in conjugate pairs. If p + iq is a root of ax² + bx + c = 0. Then p – iq is also a root of the equation.

→ Nature of the roots of ax² + bx + c = 0

Nature of the Roots	Condition
Imagine	b² – 4ac < 0
Equal	b² – 4ac = 0
Real	b² – 4ac ≥ 0
Real and different	b² – 4ac > 0
Rational	b² – 4ac is a perfect square a, b, c being rational
Equal in magnitude and opposite in sign	b = 0
Reciprocal to each other	c = a
Both positive	b has a sign opposite to that of a and c
Both negative	a, b, c all have same sign
Opposite sign	a, c are of opposite sign

→ Two equations a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have exactly the same roots if [latex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/latex]

→ The equations a₁x² + b₁x + c₁ = 0, a2x² + b₂x + c₂ = 0 have a common root, if (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁) and the common root is [latex]\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}[/latex] if a₁b₂ ≠ a₂b₁

→ If f(x) = 0 is a quadratic equation, then the equation whose roots are

The reciprocals of the roots of f(x) = 0 is f[latex]\left(\frac{1}{x}\right)[/latex] = 0
The roots of f(x) = 0, each ‘increased’ by k is f(x – k) = 0
The roots of f(x) = 0, each ‘diminished’ by k is f(x + k) = 0
The roots of f(x) = 0 with sign changed is f(-x) = 0
The roots of f(x) = 0 each multiplied by k(≠0) is f[latex]\left(\frac{x}{k}\right)[/latex] = 0

→ Sign of the expression ax² + bx + c = 0

The sign of the expression ax² + bx + c is same as that of ‘a’ for all values of x if b² – 4ac ≤ 0 i.e. if the roots of ax² + bx + c = 0 are imaginary or equal.
If the roots of the equation ax² + bx + c = 0 are real and different i.e b² – 4ac > 0, the sign of the expression is same as that of ‘a’ if x does not lie between the two roots of the equation and opposite to that of ‘a’ if x lies between the roots of the equation.

→ The expression ax² + bx + c is positive for all real values of x if b² – 4ac < 0 and a > 0.

→ The expression ax² + bx + c has a maximum value when ‘a’ is negative and x = -[latex]\frac{\mathrm{b}}{2 \mathrm{a}}[/latex]. Maximum value of the expression = [latex]\frac{4 a c-b^{2}}{4 a}[/latex]

→ The expression ax² + bx + c has a maximum value when ‘a’ is positive and x = -[latex]\frac{\mathrm{b}}{2 \mathrm{a}}[/latex]. Minimum value of the expression = [latex]\frac{4 a c-b^{2}}{4 a}[/latex]

Theorem 1:
If the roots of ax² + bx + c = 0 are imaginary, then for x ∈ R , ax² + bx + c and a have the same sign.
Proof:
The root are imaginary
b² – 4ac < 0 4ac – b² > 0
[latex]\frac{a x^{2}+b x+c}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}[/latex]
∴ For x ∈ R, ax² + bx + c = 0 and a have the same sign.

Theorem 2.
If the roots of ax² + bx + c = 0 are real and equal to α = [latex]\frac{-b}{2 a}[/latex], then α ≠ x ∈ R ax² + bx + c and a will have same sign.
Proof:
The roots of ax² + bx + c = 0 are real and equal
⇒ b² = 4ac ⇒ 4ac – b² = 0
[latex]\frac{a x^{2}+b x+c}{a}[/latex] = x + [latex]\frac{b}{a}[/latex]x + [latex]\frac{c}{a}[/latex]
= [latex]\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}[/latex]
= [latex]\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}[/latex]
= [latex]\left(x+\frac{b}{2 a}\right)^{2}[/latex] > 0 for x ≠ [latex]\frac{-b}{2 a}[/latex] = α
For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Theorem 3.
Let be the real roots of ax² + bx + c = 0 and α < β. Then
(i) x ∈ R, α < x< β ax² + bx + c and a have the opposite signs
(ii) x ∈ R, x < α or x> β ax² + bx + c and a have the same sign.
Proof:
α, β are the roots of ax² + bx + c = 0
Therefore, ax² + bx + c = a(x – α)(x – β)
[latex]\frac{a x^{2}+b x+c}{a}[/latex] = (x – α)(x – β)

(i) Suppose x ∈ R, α < x < β
⇒ x < α < β then x – α < 0, x – β < 0 ⇒ (x – α)(x – β) > 0 ⇒ [latex]\frac{a x^{2}+b x+c}{a}[/latex] > 0
⇒ ax² + bx + c, a have a same sign

(ii) Suppose x ∈ R, x > β, x > β > α then x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0 ⇒ [latex]\frac{a x^{2}+b x+c}{a}[/latex] > 0
⇒ ax² + bx + c, a have same sign
∴ x ∈ R, x < α or x > β ⇒ ax² + bx + c and a have the same sign.