Algebra Math Help

Quadratic Formula : Lucid Explanation Of Its Derivation And Application In Solving Problems

Derivation of Quadratic formula:

To solve ax^2 + bx + c = 0 where a ( not 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 + (b/a)x = -c/a

or x^2 + 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

x^2 + 2x(b/2a) + (b/2a)^2 = -c/a + (b/2a)^2

or (x + b/2a)^2 = b^2/4a^2 – c/a = ( b^2 – 4ac)/(4a^2)

or (x + b/2a) = ±SquareRoot{( b^2 – 4ac)/(4a^2)} = ±SquareRot( b^2 – 4ac)/2a

or x = -b/2a ±SquareRoot(b^2 – 4ac)/2a

or x = {-b ±SquareRoot(b^2 – 4ac)}/2a

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 ±SquareRoot(b^2 – 4ac)}/2a

= [ (-1) ±SquareRoot{(1)^2 - 4(1)(-42)}]/2(1)

= [ (-1) ±SquareRoot{1 + 168}]/2(1) = [ (-1) ±SquareRoot{169}]/2(1) = [(-1) ± 13]/2(1)

= (-1 + 13)/2, (-1 – 13)/2 = 12/2, -14/2 = 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) ±SquareRoot(b^2 – 4ac)}/2a

= [ (-6) ±SquareRoot{(6)^2 - 4(5)(-8)}]/2(5)

= [ (-6) ±SquareRoot{36 + 160}]/10 = [ (-6) ±SquareRoot{196}]/10 = [(-6) ?} 14]/10

= (-6 + 14)/10, (-6 – 14)/10 = 8/10, -20/10 = 4/5, -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) ±SquareRoot(b^2 – 4ac)}/2a

= [(-3) ±SquareRoot{(3)^2 - 4(2)(-3)}]/2(2)

= [(-3) ±SquareRoot{9 + 24}]/4 = [-3 ?} ?ã(33)]/4 Ans.

II To find the nature of the roots :

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

alpha = {-b + SquareRoot(b^2 – 4ac)}/2a and beta = {-b – SquareRoot(b^2 – 4ac)}/2a.

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

Three cases arise depending on the value of

Delta (= b^2 – 4ac) is zero or positive or negative.

(i) If Delta ( = b^2 – 4ac) = 0, then alpha = -b/2a and beta = -b/2a

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

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

(ii) If Delta ( = b^2 – 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 ( = b^2 – 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, 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 = Delta = 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 = Delta = 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 = Delta = 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 = Delta = 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

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

Then By Quadratic Formula

alpha = {-b + SquareRoot(b^2 – 4ac)}/2a and beta = {-b – SquareRoot(b^2 – 4ac)}/2a

Sum of the roots = alpha + beta

= {-b + SquareRoot(b^2 – 4ac)}/2a + {-b – SquareRoot(b^2 – 4ac)}/2a

= {-b + SquareRoot(b^2 – 4ac) -b – SquareRoot(b^2 – 4ac)}/2a

= {-2b}/2a = -b/a = -{(coefficient of x)/(coefficient of x^2)}.

Product of the roots = (alpha)(beta)

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

= [{-b + SquareRoot(b^2 - 4ac)}][{-b - SquareRoot(b^2 - 4ac)}]/(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 = alphabeta

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

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

= c/a = (constant term)/(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 = -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 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 = -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 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 = -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 x^2 – (alpha + beta)x + alphabeta = 0.

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

x^2 – (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

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 + 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) = 5^2 – SquareRoot7)^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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