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Quadratic Formula : Derivation and Application to Problems

Lesson Plans > Mathematics > Algebra > Factoring > Quadratic Formula
 

Quadratic Formula : Derivation and Application to Problems

Derivation of Quadratic formula:

To solve ax2 + bx + c = 0 where a ( not 0 ), b, c are constants which can take real number values.

ax2 + bx + c = 0

or ax2 + bx = -c

Dividing by 'a' on both sides, we get

x2 + (b⁄a)x = -c⁄a

or 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

or (x + b⁄2a)2 = b2⁄4a2 - c⁄a = ( b2 - 4ac)⁄(4a2)

or (x + b⁄2a) = PlusMinusSquareRoot{( b2 - 4ac)⁄(4a2)} = PlusMinusSquareRot( b2 - 4ac)⁄2a

or x = -b⁄2a PlusMinusSquareRoot(b2 - 4ac)⁄2a

or x = {-b PlusMinusSquareRoot(b2 - 4ac)}⁄2a

This is the Quadratic Formula. (Derived.)

I Applying Quadratic Formula in Finding the roots :

Example I(1) :

Solve x2 + x - 42 = 0 using Quadratic Formula.

Comparing this equation with ax2 + bx + c = 0, we get

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

Applying Quadratic Formula here, we get

x = {-b PlusMinusSquareRoot(b2 - 4ac)}⁄2a

= [ (-1) PlusMinusSquareRoot{(1)2 - 4(1)(-42)}]⁄2(1)

= [ (-1) PlusMinusSquareRoot{1 + 168}]⁄2(1) = [ (-1) PlusMinusSquareRoot{169}]⁄2(1) = [(-1) PlusMinus 13]⁄2(1)

= (-1 + 13)⁄2, (-1 - 13)⁄2 = 12⁄2, -14⁄2 = 6, -7 Ans.

Example I(2) :

Solve 8 - 5x2 - 6x = 0 using Quadratic Formula

Multiplying the given equation by -1, we get

5x2 + 6x - 8 = 0(-1) = 0

Comparing this equation with ax2 + bx + c = 0, we get

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

Applying Quadratic Formula here, we get

x = {(-b) PlusMinusSquareRoot(b2 - 4ac)}⁄2a

= [ (-6) PlusMinusSquareRoot{(6)2 - 4(5)(-8)}]⁄2(5)

= [ (-6) PlusMinusSquareRoot{36 + 160}]⁄10 = [ (-6) PlusMinusSquareRoot{196}]⁄10 = [(-6) } 14]⁄10

= (-6 + 14)⁄10, (-6 - 14)⁄10 = 8⁄10, -20⁄10 = 4⁄5, -2 Ans.

Example I(3) :

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

Applying Quadratic Formula here, we get

x = {(-b) PlusMinusSquareRoot(b2 - 4ac)}⁄2a

= [(-3) PlusMinusSquareRoot{(3)2 - 4(2)(-3)}]⁄2(2)

= [(-3) PlusMinusSquareRoot{9 + 24}]⁄4 = [-3 } ã(33)]⁄4 Ans.

II To find the nature of the roots :

By Quadratic Formula, the roots of ax2 + bx + c = 0 are

alpha = {-b + SquareRoot(b2 - 4ac)}⁄2a and beta = {-b - SquareRoot(b2 - 4ac)}⁄2a.

Let (b2 - 4ac) be denoted by Delta (called Delta).

Then alpha = (-b + SquareRootDelta)⁄2a and beta = (-b - SquareRootDelta)⁄2a.

The nature of the roots (alpha and beta) depends on Delta.

Delta ( = b2 - 4ac) is called the DISCRIMINANT of ax2 + bx + c = 0.

Three cases arise depending on the value of

Delta (= b2 - 4ac) is zero or positive or negative.

(i) If Delta ( = b2 - 4ac) = 0, then alpha = -b⁄2a and beta = -b⁄2a

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

Thus ax2 + bx + c = 0 has real and equal roots, if Delta = 0.

(ii) If Delta ( = b2 - 4ac) > 0, the roots are real and distinct.

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

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

(iii) If Delta ( = b2 - 4ac) < 0, SquareRootDelta is not real.

It is called an imaginary number.

i.e. alpha, beta are imaginary when Delta is negative.

When Delta is negative, the roots are imaginary.

Example II(1) :

Find the nature of the roots of the equation, 5x2 - 2x - 7 = 0.

Solution :

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 = Delta = 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.

Example II(2) :

Find the nature of the roots of the equation, 9x2 + 24x + 16 = 0.

Solution :

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 = Delta = 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.

Example II(3) :

Find the nature of the roots of the equation, x2 + 6x - 5 = 0.

Solution :

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 = Delta = 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.

Example II(4) :

Find the nature of the roots of the equation, x2 - x + 5 = 0.

Solution :

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 = Delta = b2 - 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 ax2 + bx + c = 0 be

alpha (called alpha) and beta (called beta).

Then By Quadratic Formula

alpha = {-b + SquareRoot(b2 - 4ac)}⁄2a and beta = {-b - SquareRoot(b2 - 4ac)}⁄2a

Sum of the roots = alpha + beta

= {-b + SquareRoot(b2 - 4ac)}⁄2a + {-b - SquareRoot(b2 - 4ac)}⁄2a

= {-b + SquareRoot(b2 - 4ac) -b - SquareRoot(b2 - 4ac)}⁄2a

= {-2b}⁄2a = -b⁄a = -{(coefficient of x)⁄(coefficient of x2)}.

Product of the roots = (alpha)(beta)

= [{-b + SquareRoot(b2 - 4ac)}⁄2a][{-b - SquareRoot(b2 - 4ac)}⁄2a]

= [{-b + SquareRoot(b2 - 4ac)}][{-b - SquareRoot(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.

Thus, Product of the roots = alphabeta

= [(-b)2 - {SquareRoot(b2 - 4ac)}2]⁄(4a2)

= [b2 - (b2 - 4ac)]⁄(4a2) = [b2 - b2 + 4ac)]⁄(4a2) = (4ac)⁄(4a2)

= c⁄a = (constant term)⁄(coefficient of x2)

Example III(1) :

Find the sum and product of the roots of the equation 3x2 + 2x + 1 = 0.

Solution :

The given equation is 3x2 + 2x + 1 = 0.

Comparing this equation with ax2 + bx + c = 0, we get

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

Sum of the roots = -b⁄a = -2⁄3.

Product of the roots = c⁄a = 1⁄3.

Example III(2) :

Find the sum and product of the roots of the equation x2 - px + pq = 0.

Solution :

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.

Sum of the roots = -b⁄a = -(-p)⁄1 = p.

Product of the roots = c⁄a = pq ⁄1 = pq.

Example III(3) :

Find the sum and product of the roots of the equation lx2 + lmx + lmn = 0.

Solution :

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.

Sum of the roots = -b⁄a = -(lm)⁄ l = -m

Product of the roots = c⁄a = lmn⁄l = mn.

IV To find the Quadratic Equation whose roots are given :

Let alpha and beta be the roots of the Quadratic Equation.

Then, we know (x - alpha)(x - beta) = 0.

or x2 - (alpha + beta)x + alphabeta = 0.

But, (alpha + beta) = sum of the roots and alphabeta = Product of the roots.

The required equation is

x2 - (sum of the roots)x + (product of the roots) = 0.

Thus, The Quadratic Equation with roots alpha and beta is

x2 - (alpha + beta)x + alphabeta = 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

x2 - (sum of the roots)x + (product of the roots) = 0.

So, The required equation is x2 - (1)x + (-6) = 0.

i.e. x2 - 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(m2)n.

We know the Quadratic Equation whose roots are given is

x2 - (sum of the roots)x + (product of the roots) = 0.

So, The required equation is x2 - m(l + n)x + l(m2)n = 0. Ans.

Example IV(3) :

Find the quadratic equation whose roots are (5 + SquareRoot7), (5 - SquareRoot7).

Solution:

The given roots are 5 + SquareRoot7, 5 - SquareRoot7.

Sum of the roots = (5 + SquareRoot7) + (5 - SquareRoot7) = 10;

Product of the roots = (5 + SquareRoot7)(5 - SquareRoot7) = 52 - SquareRoot7)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.

So, The required equation is x2 - (10)x + (18) = 0.

i.e. x2 - 10x + 18 = 0 Ans.

For more about Quadratic Formula, go to,

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

Lesson by kvln

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