QUADRATIC FORMULA - DERIVATION, NATURE OF ROOTS, SOLVED EXAMPLES, EXERCISE

Please study
Solving Quadratic Equations by Factoring before Quadratic Formula,
if you have not already done so.

There, we discussed about Quadratic Polynomial
or Quadratic Expression, Quadratic Equation,
Methods to Solve a Quadratic Equation and the first
method of Solving Quadratic Equations by Factoring.

Here we discuss the second method.

We derive the Formula used.

We also deal with Relation between roots and coefficients,
Nature of the roots, finding Quadratic Equation whose roots are given.

We apply all the ideas discussed, to Problems.

We explain a number of solved examples in a lucid way.

We also present a number of problems for practice in the exercise.

Solving a Quadratic Equation by Quadratic Formula

Derivation of Quadratic formula :

To solve ax² + bx + c = 0 where a ( ≠ 0 ), b, c are constants which can take real number values.

ax² + bx + c = 0 ⇒ ax² + bx = -c
Dividing by a on both sides, we get
x² + (b⁄a)x = -c⁄a
⇒ x² + 2x(b⁄2a) = -c⁄a .........(i)
The L.H.S. of equation(i) has (first term)² and 2(first term)(second term) terms where fist term = x and second term = (b⁄2a).
If we add (second term)² { = (b⁄2a)² }, the L.H.S. of equation(i) becomes a perfect square.
Adding (b⁄2a)² to both sides of equation(i), we get
x² + 2x(b⁄2a) + (b⁄2a)² = -c⁄a + (b⁄2a)²
⇒ (x + b⁄2a)² = b²⁄4a² - c⁄a = ( b² - 4ac)⁄(4a²)
⇒ (x + b⁄2a) = ±√{( b² - 4ac)⁄(4a²)} = ±√( b² - 4ac)⁄2a
⇒ x = -b⁄2a ± √(b² - 4ac)⁄2a
⇒ x = {-b ± √(b² - 4ac)}⁄2a
This is the Quadratic Formula.

ax² + bx + c = 0 ⇒ x = {-b ± √(b² - 4ac)}⁄2a

Relation between roots and coefficients of a Quadratic Equation

Let the roots of ax² + bx + c = 0 be α (called alpha) and β (called beta).
Then By Quadratic Formula
α = {-b + √(b² - 4ac)}⁄2a and β = {-b - √(b² - 4ac)}⁄2a

Sum of the roots = α + β = {-b + √(b² - 4ac)}⁄2a + {-b - √(b² - 4ac)}⁄2a
= {-b + √(b² - 4ac) -b - √(b² - 4ac)}⁄2a
= {-2b}⁄2a = -b⁄a

Sum of the roots = α + β = -b⁄a = -{(coefficient of x)⁄(coefficient of x²)}

Product of the roots = α x β = {-b + √(b² - 4ac)}⁄2a x {-b - √(b² - 4ac)}⁄2a
= [{-b + √(b² - 4ac)} x {-b - √(b² - 4ac)}]⁄(4a²)
The Numerator is product of sum and difference of two terms which we know is equal to the difference of the squares of the two terms.
∴ Product of the roots = αβ = [(-b)² - {√(b² - 4ac)}²]⁄(4a²)
= [b² - (b² - 4ac)]⁄(4a²) = [b² - b² + 4ac)]⁄(4a²) = (4ac)⁄(4a²)
= c⁄a

Product of the roots = αβ = c⁄a = (constant term)⁄(coefficient of x²)}

Nature of the roots of a Quadratic Equation :

By Quadratic Formula, the roots of ax² + bx + c = 0
are α = {-b + √(b² - 4ac)}⁄2a and β = {-b - √(b² - 4ac)}⁄2a
Let (b² - 4ac) be denoted by Δ (called Delta).
Then α = (-b + √Δ)⁄2a and β = (-b - √Δ)⁄2a
The nature of the roots (α and β) depends on Δ
Δ ( = b² - 4ac) is called the DISCRIMINANT of ax² + bx + c = 0.
Three cases arise depending on the value of Δ (= b² - 4ac) is zero or positive or negative.
(i) If Δ ( = b² - 4ac) = 0, then α = -b⁄2a and β = -b⁄2a i.e. the two roots are real and equal.
Thus ax² + bx + c = 0 has real and equal roots, if Δ = 0
(ii) If Δ ( = b² - 4ac) > 0, the roots are real and distinct.
(ii) (a) If Δ ( = b² - 4ac) is a perfect square,
the roots are rational.
(ii) (b) otherwise, the roots are irrational.
(iii) If Δ ( = b² - 4ac) < 0, √Δ is not real.It is called an imaginary number.
∴ α, β are imaginary when Δ is negative.
When Δ is negative, the roots are imaginary.

The roots of ax² + bx + c = 0
(i) are real and equal if Δ ( = b² - 4ac) = 0
(ii) are real and distinct if Δ ( = b² - 4ac) > 0
(ii) (a) If Δ ( = b² - 4ac) is a perfect square,
the roots are rational.
(ii) (b) otherwise, the roots are irrational.
(iii) are imaginary if Δ ( = b² - 4ac) < 0

To find the Quadratic Equation whose roots are given :

Let α and β be the roots of the Quadratic Equation.

Example 1 : Quadratic Formula

Solve the following equations using Quadratic Formula

1. x² + x - 42 = 0
2. 3x² + 2x - 8 = 0
3. 8 - 5x² - 6x = 0
4. 6x² - 13x - 63 = 0
5. 2x² + 3x - 3 = 0

(ii) To solve 3x² + 2x - 8 = 0 using Quadratic Formula
Comparing this equation with ax² + bx + c = 0, we get
a = 3, b = 2 and c = -8
We know by Quadratic Formula, x = {(-b) ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b² - 4ac)}⁄2a
= [ (-2) ± √{(2)² - 4(3)(-8)}]⁄2(3)
= [ (-2) ± √{4 + 96}]⁄6 = [ (-2) ± √{100}]⁄6 = [(-2) ± 10]⁄6
= (-2 + 10)⁄6, (-2 - 10)⁄6 = 8⁄6, -12⁄6 = 4⁄3, -2 Ans.

(iii) To solve 8 - 5x² - 6x = 0 using Quadratic Formula
Multiplying the given equation by -1, we get
5x² + 6x - 8 = 0 x -1 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 5, b = 6 and c = -8
We know by Quadratic Formula, x = {(-b) ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b² - 4ac)}⁄2a
= [ (-6) ± √{(6)² - 4(5)(-8)}]⁄2(5)
= [ (-6) ± √{36 + 160}]⁄10 = [ (-6) ± √{196}]⁄10 = [(-6) ± 14]⁄10
= (-6 + 14)⁄10, (-6 - 14)⁄10 = 8⁄10, -20⁄10 = 4⁄5, -2 Ans.

(iv) To solve 6x² - 13x - 63 = 0 using Quadratic Formula
Comparing this equation with ax² + bx + c = 0, we get
a = 6, b = -13 and c = -63
We know by Quadratic Formula, x = {(-b) ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b² - 4ac)}⁄2a
= [ -(-13) ± √{(-13)² - 4(6)(-63)}]⁄2(6)
= [ (+13) ± √{169 + 1512}]⁄12 = [ (13) ± √{1681}]⁄12 = [(13) ± 41]⁄12
= (13 + 41)⁄12, (13 - 41)⁄12 = 54⁄12, -28⁄12 = 9⁄2, -7⁄3 Ans.

(v) To solve 2x² + 3x - 3 = 0 using Quadratic Formula
Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = 3 and c = -3
We know by Quadratic Formula, x = {(-b) ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b² - 4ac)}⁄2a
= [ (-3) ± √{(3)² - 4(2)(-3)}]⁄2(2)
= [ (-3) ± √{9 + 24}]⁄4 = [-3 ± √(33)]⁄4 Ans.

Example 2 : Quadratic Formula

Solve the following equation using Quadratic Formula
2x⁴ - 9x³ + 14x² - 9x + 2 = 0


Solution of Example 1 of Quadratic Formula :

The given equation in descending powers of x is
2x⁴ - 9x³ + 14x² - 9x + 2 = 0
This is an equation of the form ax⁴ + bx³ + cx² + bx + a = 0
in which the coefficients of terms equidistant from first and last are equal, is called a RECIPROCAL EQUATION. By dividing this reciprocal equation with x² and with a proper substitution, the reciprocal equation can be reduced to a Quadratic Equation.
Dividing the equation by x², we get
2x² - 9x + 14 - 9x^-1 + 2x^-2 = 0
⇒ 2x² + 2⁄x² - 9x - 9⁄x + 14 = 0
⇒ 2(x² + 1⁄x²) -9(x + 1⁄x) + 14 = 0
Putting x + 1⁄x = t, x² + 1⁄x² = (x + 1⁄x)² - 2 = t² - 2
∴ the given equation becomes 2(t² - 2) -9t + 14 = 0
⇒ 2t² - 4 - 9t + 14 = 0 ⇒ 2t² - 9t + 10 = 0
Comparing this equation with at² + bt + c = 0, we get
a = 2, b = -9 and c = 10
We know by Quadratic Formula, t = {-b ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
t = {-b ± √(b² - 4ac)}⁄2a
= [-(-9) ± √{(-9)² - 4(2)(10)}]⁄2(2)
= [ (+9) ± √{81 - 80}]⁄4 = [ (9) ± √{1}]⁄4 = [ (9) ± 1]⁄4
= (9 + 1)⁄4, (9 - 1)⁄4 = 10⁄4, 8⁄4 = 5⁄2, 2

But t = x + 1⁄x

Taking t = 5⁄2,
t = 5⁄2 ⇒ x + 1⁄x = 5⁄2
Multiplying both sides with 2x, we get
2x² + 2 = 5x ⇒ 2x² - 5x + 2 = 0 Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -5 and c = 2
We know by Quadratic Formula, x = {-b ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b² - 4ac)}⁄2a
= [ -(-5) ± √{(-5)² - 4(2)(2)}]⁄2(2)
= [ (+5) ± √{25 - 16}]⁄4 = [ (5) ± √{9}]⁄4 = [ (5) ± 3]⁄4
= (5 + 3)⁄4, (5 - 3)⁄4 = 8⁄4, 2⁄4 = 2, 1⁄2

Taking t = 2,
t = 2 ⇒ x + 1⁄x = 2
Multiplying both sides with x, we get
x² + 1 = 2x ⇒ x² - 2x + 1 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -2 and c = 1
We know by Quadratic Formula, x = {-b ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b² - 4ac)}⁄2a
= [ -(-2) ± √{(-2)² - 4(1)(1)}]⁄2(1)
= [ (+2) ± √{4 - 4}]⁄2 = [ (2) ± √{0}]⁄2 = [ (2) ± 0]⁄2
= (2 + 0)⁄2, (2 - 0)⁄2 = 2⁄2, 2⁄2 = 1, 1

Thus the solution of the given equation is x = { 1, 1, 2, 1⁄2 } Ans.

Example 3 : Quadratic Formula

Solve the following equation using Quadratic Formula
√(x + 1) + √(x - 2) = √(x + 3)

Solution of Example 1 of Quadratic Formula :

The given equation is √(x + 1) + √(x - 2) = √(x + 3)
Squaring both sides, we get
{√(x + 1) + √(x - 2)}² = {√(x + 3)}²
⇒ {√(x + 1)}² + {√(x - 2)}² + 2{√(x + 1)}{√(x - 2)} = (x + 3)

⇒ (x + 1) + (x - 2) + 2[√{(x + 1)(x - 2)] = (x + 3)

⇒ (2x - 1) - (x + 3) = -2[√{(x + 1)(x - 2)] ⇒ 2x - 1 - x - 3 = -2[√{(x + 1)(x - 2)] ⇒ x - 4 = -2[√{(x + 1)(x - 2)]
Squaring both sides, we get
(x - 4)² = 4(x + 1)(x - 2) ⇒ x² - 8x + 16 = 4(x² - x - 2) = 4x² - 4x - 8
⇒ x² - 4x² - 8x + 4x + 16 + 8 = 0 ⇒ -3x² - 4x + 24 = 0
⇒ 3x² + 4x - 24 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 3, b = 4 and c = -24
We know by Quadratic Formula, x = {(-b) ± √(b² - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b² - 4ac)}⁄2a
= [ -(4) ± √{(4)² - 4(3)(-24)}]⁄2(3)
= [ (-4) ± √{16 + 288}]⁄6 = [-4 ± √(304)]⁄6 = [-4 ± √{4(76)}]⁄6
= [-4 ± 2{√(76)}]⁄6 = [-2 ± {√(76)}]⁄3. Ans.

Example 4 : Quadratic Formula

Find the sum and product of the roots of the equations given below.

1. 3x² + 2x + 1 = 0
2. x² - px + pq = 0
3. lx² + lmx + lmn = 0

(ii) The given equation is x² - px + pq = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -p and c = pq
We know, Sum of the roots = -b⁄a = -(-p)⁄1 = p Ans.
Product of the roots = c⁄a = pq ⁄1 = pq Ans.

(iii) The given equation is lx² + lmx + lmn = 0
Comparing this equation with ax² + bx + c = 0, we get
a = l, b = lm and c = lmn
We know, Sum of the roots = -b⁄a = -(lm)⁄ l = -m Ans.
Product of the roots = c⁄a = lmn⁄l = mn Ans.

Example 5 : Quadratic Formula

If α, β are the roots of x² - px + q = 0, then find the value of

1. α³ + β³
2. (α²⁄β) + (β²⁄α)
3. (1⁄α³) + (1⁄β³)
4. (α⁄β) + (β⁄α)

(i) To find the value of α³ + β³

We know from Algebra Formula 6 in Algebra Formulas :

Cube of Sum of Two Terms:

(a + b)³ = a³ + 3ab(a + b) + b³

(ii) To find the value of (α²⁄β) + (β²⁄α)

Let A = (α²⁄β) + (β²⁄α)
Multiplying both sides with αβ, we get
αβA = αβ(α²⁄β) + αβ(β²⁄α) = α³ + β³
Using the values of α³ + β³ (= p³ - 3pq) and αβ (= q) here, we get
qA = p³ - 3pq ⇒ A = (p³ - 3pq)⁄q ⇒ (α²⁄β) + (β²⁄α) = (p³ - 3pq)⁄q. Ans.

(iii) To find the value of (1⁄α³) + (1⁄β³)

Let A = (1⁄α³) + (1⁄β³)
Multiplying both sides with α³β³, we get
α³β³A = α³β³(1⁄α³) + α³β³ (1⁄β³)
(αβ)³A = β³ + α³ = α³ + β³
Using the values of α³ + β³ (= p³ - 3pq) and αβ (= q) here, we get
q³A = p³ - 3pq ⇒ A = (p³ - 3pq)⁄q³ ⇒ (1⁄α³) + (1⁄β³) = (p³ - 3pq)⁄q³. Ans.

(iv) To find the value of (α⁄β) + (β⁄α)
Let A = (α⁄β) + (β⁄α)
Multiplying both sides with αβ, we get
αβA = αβ(α⁄β) + αβ(β⁄α) = α² + β² = (α + β)² - 2αβ
Using the values of α + β (= p) and αβ (= q) here, we get
qA = (p)² - 2q ⇒ A = (p² - 2q)⁄q
⇒ (α⁄β) + (β⁄α) = (p² - 2q)⁄q. Ans.

Example 6 : Quadratic Formula

Find the nature of the roots of the equations,

1. 5x² - 2x - 7 = 0
2. 9x² + 24x + 16 = 0
3. x² + 6x - 5 = 0
4. x² - x + 5 = 0

(ii) The given equation is 9x² + 24x + 16 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 9, b = 24 and c = 16
Discriminant = Δ = b² - 4ac = (24)² - 4(9)(16) = 576 - 576 = 0
Since the Discriminant is zero,
the roots of the given equation
are real and equal. Ans.

(iii) The given equation is x² + 6x - 5 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = 6 and c = -5
Discriminant = Δ = b² - 4ac = (6)² - 4(1)(-5) = 36 + 20 = 56
Since the Discriminant is positive and is not a perfect square,
the roots of the given equation
are real, distinct and irrational. Ans.

(iv) The given equation is x² - x + 5 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -1 and c = 5
Discriminant = Δ = b² - 4ac = (-1)² - 4(1)(5) = 1 - 20 = -19
Since the Discriminant is negative,
the roots of the given equation
are imaginary. Ans.

Example 7 : Quadratic Formula

Find the quadratic equation whose roots are

1. 3, -2
2. 4⁄3, 1⁄3
3. lm, mn
4. (5 + √7), (5 - √7)

(ii) The given roots are 4⁄3, 1⁄3
Sum of the roots = 4⁄3 + 1⁄3 = 5⁄3; Product of the roots = 4⁄3 x 1⁄3 = 4⁄9
We know the Quadratic Equation whose roots are given is
x² - (sum of the roots)x + (product of the roots) = 0
∴ The required equation is x² - (5⁄3)x + (4⁄9) = 0
Multiplying both sides with 9, we get
9x² -15x + 4 = 0 Ans.

(iii) The given roots are lm, mn
Sum of the roots = lm + mn = m(l + n); Product of the roots = lm x mn = lm²n
We know the Quadratic Equation whose roots are given is
x² - (sum of the roots)x + (product of the roots) = 0
∴ The required equation is x² - m(l + n)x + lm²n = 0. Ans.

(iv) The given roots are 5 + √7, 5 - √7
Sum of the roots = (5 + √7) + (5 - √7) = 10;
Product of the roots = (5 + √7)x(5 - √7) = 5² - (√7)² = 25 - 7 = 18
We know the Quadratic Equation whose roots are given is
x² - (sum of the roots)x + (product of the roots) = 0
∴ The required equation is x² - (10)x + (18) = 0
i.e x² - 10x + 18 = 0 Ans.

Example 8 : Quadratic Formula

If one root of x² - 5x + k = 0 is 2, find the value of k and the other root.

Solution of Example 8 of Quadratic Formula :

The given equation is x² - 5x + k = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -5 and c = k
By data one root is 2. Let the other root be β
Then 2 + β = -b⁄a = -(-5)⁄1 = 5; ..........(i)
2(β) = c⁄a = k⁄1 = k ⇒ β = k⁄2 ......(ii)
Using (ii) in (i), we get 2 + k⁄2 = 5 ⇒ k⁄2 = 5 - 2 = 3
⇒ k = 3 x 2 = 6. Ans.
The other root = β = k⁄2 = 6⁄2 = 3. Ans.

Example 9 : Quadratic Formula

A root of px² + qx + r = 0 is thrice the other root. Show that 3q² = 16pr.

Solution of Example 9 of Quadratic Formula :

The given equation is px² + qx + r = 0
Comparing this equation with ax² + bx + c = 0, we get
a = p, b = q and c = r
By data one root is thrice the other root.
⇒ If one root is α, then the other root is 3α
α + 3α = -b⁄a = -q⁄p
⇒ 4α = -q⁄p ⇒ α = -q⁄4p......(i)
α x 3α = c⁄a ⇒ 3α² = r⁄p......(ii)
Using (i) in (ii), we get
3(-q⁄4p)² = r⁄p ⇒ 3q²⁄16p² = r⁄p
Multiplying both sides with 16p², we get
3q² = (16p²)(r⁄p) = 16pr (Proved.)

Example 10 : Quadratic Formula

For what values of k are the roots of kx² + (k - 1)x + (k -1) = 0 equal ? For those values of k, find the equal roots.

Solution of Example 10 of Quadratic Formula :

The given equation is kx² + (k - 1)x + (k -1) = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = (k - 1) and c = (k -1)
Discriminant = Δ = b² - 4ac = (k - 1)² - 4k(k -1) = (k -1)(k - 1 - 4k)
= (k -1)(- 1 - 3k)
For roots to be equal, Discriminant = 0
⇒ (k -1)(- 1 - 3k) = 0 ⇒ (k -1) = 0 or (- 1 - 3k) = 0
⇒ k = 1 or -3k = 1 ⇒ k = 1 or k = -1⁄3 ⇒ k = 1, -1⁄3
roots = (-b ± 0)⁄2a = {-(k - 1)± 0}⁄2k = (1 - k)⁄2k
For k = 1, the equal roots are 0, 0.
For k = -1⁄3, the value of the equal roots is {1 - (-1⁄3)}⁄2(-1⁄3) = -2
Thus, for k = 1, the equal roots are 0, 0
and for k = -1⁄3, the equal roots are -2, -2. Ans.

Exercise 2 : Quadratic Formula

1. Solve the following equations using Quadratic Formula
   
   1. x² - 4x - 12 = 0
   2. x² + 16x + 48 = 0
   3. 10x² - 7x - 12 = 0
      
   4. 16x - 15 - 4x² = 0
   5. 12x² + 3x - 99 = 0
2. Solve the following equation using Quadratic Formula
   6x⁴ - 35x³ + 62x² - 35x + 6 = 0
   
3. Solve the following equations using Quadratic Formula
   √(x + 8) - √(x + 3) = √x
4. Find the sum and product of the roots of the equations given below.
   
   1. px² - rx + q = 0
   2. x² - px + q = 0
   3. 9x² + 4x - 11 = 0
5. If α, β are the roots of x² - (k + 1)x + (k² + k + 1)⁄2 = 0, then show that α² + β² = k
6. Find the nature of the roots of the equations,
   
   1. x² - 5x + 6 = 0
   2. x² - 16x + 64 = 0
   3. x² + 2x - 1 = 0
   4. x² + 4x + 5 = 0
7. Find the quadratic equation whose roots are
   
   1. -2, -4
   2. 3 + √2, 3- √2
   3. (l - m), (l + m)
   4. 1/2, 3/2
8. If one root of x² -(p - 1)x + 10 = 0 is 5, then find the value of p and the second root.
9. For what values of k does the equation
   (k - 2)x² + 2(2k - 3)x + (5k - 6) = 0 have equal roots?
   For those values of k, find those equal roots.
10. If α, β are the roots of px² + qx + r = 0, find the quadratic equation whose roots are(α + 1⁄β), (β + 1⁄α)

For Answers See at the bottom of the Page.

Answers to Exercise : Quadratic Formula

1. 1. 6, -2
   2. -4, -12
   3. -4⁄5, 3⁄2
   4. 5⁄2, 3⁄2
   5. 11⁄4, -3
2. 2, 3, 1⁄2, 1⁄3
3. 1, -25⁄3
4. 1. r⁄q, q⁄p
   2. p, q
   3. -4⁄9, -11⁄9
5. To be proved.
6. 1. real, distinct and rational
   2. real, equal
   3. real, distinct and irrational
   4. imaginary
7. 1. x² + 6x + 8 = 0
   2. x² - 6x + 7 = 0
   3. x² - 2lx + (l² - m²) = 0
   4. 4x² - 8x + 3 = 0
      
8. p = 8, second root = 2
9. k = 1, roots = -1,-1; k = 3, roots = -3,-3;
   
10. x² + (q⁄p + q⁄r)x + (p⁄r + r⁄p + 2) = 0