We explain lucidly, the derivation of Quadratic formula and applying it in finding the roots, Relation between roots and coefficients, Nature of the roots, finding Quadratic Equation whose roots are given, with Examples.

Derivation of Quadratic formula:

To solve ax^2 + bx + c = 0 where a ( ` 0 ), b, c are constants which can take real number values.

ax^2 + bx + c = 0

or ax^2 + bx = -c

Dividing by 'a' on both sides, we get

x^2 + (bDa)x = -cDa

or x^2 + 2x(bD2a) = -cDa .........(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 = (bD2a).

If we add (second term)^2 {= (bD2a)^2}, the L.H.S. of equation(i) becomes a perfect square.

Adding (bD2a)^2 to both sides of equation(i), we get

x^2 + 2x(bD2a) + (bD2a)^2 = -cDa + (bD2a)^2

or (x + bD2a)^2 = b^2D4a^2 - cDa = ( b^2 - 4ac)D(4a^2)

or (x + bD2a) = ±{( b^2 - 4ac)D(4a^2)} = ±( b^2 - 4ac)D2a

or x = -bD2a ± (b^2 - 4ac)D2a

or x = {-b ± (b^2 - 4ac)}D2a

This is the Quadratic Formula. (Derived.)

I Applying Quadratic Formula in Finding the roots :

Example I(1) :

Solve x^2 + x - 42 = 0 using Quadratic Formula.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = 1 and c = -42

Applying Quadratic Formula here, we get

x = {-b ± (b^2 - 4ac)}D2a

= [ (-1) ± {(1)^2 - 4(1)(-42)}]D2(1)

= [ (-1) ± {1 + 168}]D2(1) = [ (-1) ± {169}]D2(1) = [(-1) ± 13]D2(1)

= (-1 + 13)D2, (-1 - 13)D2 = 12D2, -14D2 = 6, -7 Ans.

Example I(2) :

Solve 8 - 5x^2 - 6x = 0 using Quadratic Formula

Multiplying the given equation by -1, we get

5x^2 + 6x - 8 = 0(-1) = 0

Comparing this equation with ax^2 + bx + c = 0, we get

a = 5, b = 6 and c = -8

Applying Quadratic Formula here, we get

x = {(-b) ± (b^2 - 4ac)}D2a

= [ (-6) ± {(6)^2 - 4(5)(-8)}]D2(5)

= [ (-6) ± {36 + 160}]D10 = [ (-6) ± {196}]D10 = [(-6) ± 14]D10

= (-6 + 14)D10, (-6 - 14)D10 = 8D10, -20D10 = 4D5, -2 Ans.

Example I(3) :

Solve 2x^2 + 3x - 3 = 0 using Quadratic Formula

Comparing this equation with ax^2 + bx + c = 0, we get

a = 2, b = 3 and c = -3

Applying Quadratic Formula here, we get

x = {(-b) ± (b^2 - 4ac)}D2a

= [(-3) ± {(3)^2 - 4(2)(-3)}]D2(2)

= [(-3) ± {9 + 24}]D4 = [-3 ± (33)]D4 Ans.

II To find the nature of the roots :

By Quadratic Formula, the roots of ax^2 + bx + c = 0 are

± = {-b + (b^2 - 4ac)}D2a and ² = {-b - (b^2 - 4ac)}D2a.

Let (b^2 - 4ac) be denoted by ” (called Delta).

Then ± = (-b + ”)D2a and ² = (-b - ”)D2a.

The nature of the roots (± and ²) depends on ”.

” ( = b^2 - 4ac) is called the DISCRIMINANT of ax^2 + bx + c = 0.

Three cases arise depending on the value of

” (= b^2 - 4ac) is zero or positive or negative.

(i) If ” ( = b^2 - 4ac) = 0, then ± = -bD2a and ² = -bD2a

i.e. the two roots are real and equal.

Thus ax^2 + bx + c = 0 has real and equal roots, if ” = 0.

(ii) If ” ( = b^2 - 4ac) > 0, the roots are real and distinct.

(ii) (a) if ” is a perfect square, the roots are rational.

(ii) (b) if ” is not a perfect square, the roots are irrational.

(iii) If ” ( = b^2 - 4ac) < 0, ” is not real.

It is called an imaginary number.

i.e. ±, ² are imaginary when ” is negative.

When ” is negative, the roots are imaginary.

Example II(1) :

Find the nature of the roots of the equation, 5x^2 - 2x - 7 = 0.

Solution :

The given equation is 5x^2 - 2x - 7 = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 5, b = -2 and c = -7.

Discriminant = ” = b^2 - 4ac = (-2)2 - 4(5)(-7) = 4 + 140 = 144 = 12^2

Since the Discriminant is positive and a perfect square,

the roots of the given equation are real, distinct and rational. Ans.

Example II(2) :

Find the nature of the roots of the equation, 9x^2 + 24x + 16 = 0.

Solution :

The given equation is 9x^2 + 24x + 16 = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 9, b = 24 and c = 16

Discriminant = ” = b^2 - 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.

Example II(3) :

Find the nature of the roots of the equation, x^2 + 6x - 5 = 0.

Solution :

The given equation is x^2 + 6x - 5 = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = 6 and c = -5.

Discriminant = ” = b^2 - 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.

Example II(4) :

Find the nature of the roots of the equation, x^2 - x + 5 = 0.

Solution :

The given equation is x^2 - x + 5 = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = -1 and c = 5.

Discriminant = ” = b^2 - 4ac = (-1)^2 - 4(1)(5) = 1 - 20 = -19.

Since the Discriminant is negative,

the roots of the given equation are imaginary. Ans.

III To find the relation between the roots and the coefficients :

Let the roots of ax^2 + bx + c = 0 be

± (called alpha) and ² (called beta).

Then By Quadratic Formula

± = {-b + (b^2 - 4ac)}D2a and ² = {-b - (b^2 - 4ac)}D2a

Sum of the roots = ± + ²

= {-b + (b^2 - 4ac)}D2a + {-b - (b^2 - 4ac)}D2a

= {-b + (b^2 - 4ac) -b - (b^2 - 4ac)}D2a

= {-2b}D2a = -bDa = -{(coefficient of x)D(coefficient of x^2)}.

Product of the roots = (±)(²)

= [{-b + (b^2 - 4ac)}D2a][{-b - (b^2 - 4ac)}D2a]

= [{-b + (b^2 - 4ac)}][{-b - (b^2 - 4ac)}]D(4a^2)

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.

Thus, Product of the roots = ±²

= [(-b)^2 - {(b^2 - 4ac)}^2]D(4a^2)

= [b^2 - (b^2 - 4ac)]D(4a^2) = [b^2 - b^2 + 4ac)]D(4a^2) = (4ac)D(4a^2)

= cDa = (constant term)D(coefficient of x^2)

Example III(1) :

Find the sum and product of the roots of the equation 3x^2 + 2x + 1 = 0.

Solution :

The given equation is 3x^2 + 2x + 1 = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 3, b = 2 and c = 1.

Sum of the roots = -bDa = -2D3.

Product of the roots = cDa = 1D3.

Example III(2) :

Find the sum and product of the roots of the equation x^2 - px + pq = 0.

Solution :

The given equation is x^2 - px + pq = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = -p and c = pq.

Sum of the roots = -bDa = -(-p)D1 = p.

Product of the roots = cDa = pq D1 = pq.

Example III(3) :

Find the sum and product of the roots of the equation lx^2 + lmx + lmn = 0.

Solution :

The given equation is lx^2 + lmx + lmn = 0.

Comparing this equation with ax^2 + bx + c = 0, we get

a = l, b = lm and c = lmn.

Sum of the roots = -bDa = -(lm)D l = -m

Product of the roots = cDa = lmnDl = mn.

IV To find the Quadratic Equation whose roots are given :

Let ± and ² be the roots of the Quadratic Equation.

Then, we know (x - ±)(x - ²) = 0.

or x^2 - (± + ²)x + ±² = 0.

But, (± + ²) = sum of the roots and ±² = Product of the roots.

The required equation is

x^2 - (sum of the roots)x + (product of the roots) = 0.

Thus, The Quadratic Equation with roots ± and ² is

x^2 - (± + ²)x + ±² = 0.

Example IV(1) :

Find the quadratic equation whose roots are 3, -2.

Solution:

The given roots are 3, -2.

Sum of the roots = 3 + (-2) = 3 - 2 = 1;

Product of the roots = 3 x (-2) = -6.

We know the Quadratic Equation whose roots are given is

x^2 - (sum of the roots)x + (product of the roots) = 0.

So, The required equation is x^2 - (1)x + (-6) = 0.

i.e. x^2 - x - 6 = 0 Ans.

Example IV(2) :

Find the quadratic equation whose roots are lm, mn.

Solution:

The given roots are lm, mn.

Sum of the roots = lm + mn = m(l + n);

Product of the roots = (lm)(mn) = l(m^2)n.

We know the Quadratic Equation whose roots are given is

x^2 - (sum of the roots)x + (product of the roots) = 0.

So, The required equation is x^2 - m(l + n)x + l(m^2)n = 0. Ans.

Example IV(3) :

Find the quadratic equation whose roots are (5 + 7), (5 - 7).

Solution:

The given roots are 5 + 7, 5 - 7.

Sum of the roots = (5 + 7) + (5 - 7) = 10;

Product of the roots = (5 + 7)(5 - 7) = 5^2 - (7)^2 = 25 - 7 = 18.

We know the Quadratic Equation whose roots are given is

x^2 - (sum of the roots)x + (product of the roots) = 0.

So, The required equation is x^2 - (10)x + (18) = 0.

i.e. x^2 - 10x + 18 = 0 Ans.

For more about Quadratic Formula, go to,

http://www.math-help-ace.com/Quadratic-Formula.html

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