Linear and Quadratic equations
What is a linear equation?
Linear equations are equations that graph as a line. They are equations that can be written as
y equals ax plus b. y= ax+b.
Notice that neither y nor x is to a power or a root.
What Is Quadratic Equation?
Quadratic equations are the polynomial equations of degree 2 in one variable of type
f(x) = ax2 + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic
equation where ‘a’ is called the leading coefficient and ‘c’ is called the absolute term of f (x).
A quadratic polynomial, when equated to zero, becomes a quadratic equation. In other terms,
a quadratic equation is a second-degree algebraic equation.
The values of x satisfying the equation are called the roots of the quadratic equation.
General from: ax2 + bx + c = 0
Here, a, b, c, ∈ R and a ≠ 0
Examples: 3x2 + x + 5 = 0, -x2 + 7x + 5 = 0, x2 + x = 0.
Quadratic Equation Formula
The solution or roots of a quadratic equation are given by the quadratic formula:
x = (α, β) = [-b ± √(b2 – 4ac)]/2a
Formulas for Solving Quadratic Equations
1. The roots of the quadratic equation: x = (-b ± √D)/2a, where D = b2 – 4ac
2. Nature of roots:
• D > 0, roots are real and distinct (unequal)
• D = 0, roots are real and equal (coincident)
• D < 0, roots are imaginary and unequal
3. The roots (α + iβ), (α – iβ) are the conjugate pair of each other.
4. Sum and product of roots: If α and β are the roots of a quadratic equation, then
• S = α+β= -b/a = -coefficient of x/coefficient of x2
• P = αβ = c/a = constant term/coefficient of x2
5. Quadratic equation in the form of roots: x2 – (α + β)x + (αβ) = 0
6. The quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have;
• One common root if (b1c2 – b2c1)/(c1a2 – c2a1) = (c1a2 – c2a1)/(a1b2 – a2b1)
• Both roots common if a1/a2 = b1/b2 = c1/c2
7. In quadratic equation ax2 + bx + c = 0 or [(x + b/2a)2 – D/4a2] = 0
• If a > 0, minimum value = 4ac – b2/4a at x = -b/2a.
• If a < 0, maximum value 4ac – b2/4a at x= -b/2a.
8. If α, β, γ are roots of cubic equation ax3 + bx2 + cx + d = 0, then α + β + γ = -b/a, αβ + βγ
+ λα = c/a, and αβγ = -d/a
9. A quadratic equation becomes an identity (a, b, c = 0) if the equation is satisfied by more
than two numbers, i.e., having more than two roots or solutions either real or complex.
Solved Examples
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.
