QUADRATIC SOLVER - NATURE OF THE ROOTS OF A QUADRATIC EQUATION, EXAMPLES, EXERCISE

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Please study
Derivation of Quadratic Formula before Quadratic Solver
if you have not already done so.

There, We presented the Derivation
Of Quadratic Formula in a lucid way.
An Example in applying the formula
is given. Another problem for practice
is also given.

That knowledge is a prerequisite here.

Here, we explain (derive) the Formulae
to decide the Nature of the roots of
a Quadratic Equation

We also present Solved Examples and
Exercise problems on application of
these Formulae.

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

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Other books just give practice.
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These provide many pages of practice that gradually
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Example 1 : Quadratic Solver

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


Solution of Example 1(i) of Quadratic Solver :

(i) The given equation is 5x2 - 2x - 7 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 5, b = -2 and c = -7
Discriminant = Δ = b2 - 4ac = (-2)2 - 4(5)(-7) = 4 + 140 = 144 = 122
Since the Discriminant is positive and a perfect square,
the roots of the given equation
are real, distinct and rational. Ans.

Solution of Example 1(ii) of Quadratic Solver :

(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.

Solution of Example 1(iii) of Quadratic Solver :

(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.

Solution of Example 1(iv) of Quadratic Solver :

(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. Great Deals on School & Homeschool Curriculum Books

Example 2 : Quadratic Solver

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 2 of Quadratic Solver :

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.

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Exercise : Quadratic Solver

Solve the following problems of Quadratic Solver :

  1. 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
  2. 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.

Progressive Learning of Math : Quadratic Solver

Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.

These also provide teachers and parents with lessons
on how to work with the child on the concepts.

The series is low to reasonably priced and include

Elementary Math curriculum

and

Algebra Curriculum.


Answers to Exercise : Quadratic Solver

Solve the following problems on Quadratic Solver

    1. real, distinct and rational
    2. real, equal
    3. real, distinct and irrational
    4. imaginary
  1. x2 + (qp + qr)x + (pr + rp + 2) = 0