QUADRATIC EQUATION

Definition: An equation in which the highest power of the variable is 2 is called QUADRATIC EQUATION

ax²+bx+c = 0, where a,b,c are constant is a general quadratic equation

FACTORIZATION OF A QUADRATIC EXPRESSION

Consider that ax²+ bx + c =a

The factor of the expression depends on the signs and values of b and c

If c is positive and a = 1

i. and are factor of c and both have sign of b

ii.The sum of and is the numerical value of b, i.e

If c is negative and a = 1

i. and are factors of c and have opposite sign

ii.The difference between and is the value of b

Example

x²+7x+12

possible factors are (2,6), (3,4), (1,12)

c is positive. The required actor adds up ;

i.e. 7 (3,4)

x² + 3x + 4x+ 12

(x+3) (x+4)

Example

x² – 11x + 28

Possible factors are: (2,14), (7,4)

c is positive. The required factors adds-up

i.e. 11 (4,7) Both factors have sign of b i.e. negative

7x² – 11x + 28 = x² – 7x– 4x + 28

= x(x–7) –4(x–7)

=(x–7) (x– 4)

Example

x² –2x – 24

Possible factor are: (2,12), (3,8), (4,6)

c is positive required factor differ by /b/

i.e. 24 (4,6) both factor have sign of b

c is negative therefore factors differs in sign.

The numerically larger having the sign of b i.e. negative

x² -2x – 24 = x² – 6x + 4x – 24

= x(x-6) +4(x-6)

= (x+4) (x-6)

Example

x² – 2x – 35

Possible factors are; (2,12), (3,8), (4,6)

c is negative, required factors differs by

i.e. 2 (5,7)

c is negative therefore factors differs in sign. The numerically larger having the sign of b i.e negative

x² – 2x – 35 = x² – 7x + 5x – 35

= x(x−7) +5(x−5)

= (x+5)(x−7)

If a 1, ax² +bx + c

The following steps should be used

1. We obtain i.e. the numerical value of the product ac ignoring the sign of the product

2. Write down all possible pair of factor of /ac/

3i. If c is positive, select two factors of ac with sum equal to

ii If c is negative, select two factors of ac that differ by

iii If c is negative, the factors have opposite sign the numerically larger having the same as b

Then, ax²+bx+c = and factorise this by the method of common

factor

Example

To factorise 2x²+7x+3

a = 2, b = 7, c = 3, = 2 ×3 = 6

possible factor of 6 = (1,6), (2,3)

Required factor = (1,6)

2x²+7x+3 = 2x²+x+6x+3

= x(2x+1) +3 (2x+1)

= (x+3) (2x+1)

Example

Factorise 8x²+2x-3

a = 8, b = 7, c = 3, = 8×3 = 24

Possible factor of 24 = (1,24), (3,8), (4,6), (2,12)

c is negative required factor differs /b/ i.e. (4,6)

c is negative sign, of therefore the sign of and are different

= 6, = −4

8x² +2x – 3 = 8x² + 6x −4x−3

= 2x(4x+3) – 1(4x+3)

= (2x−1)(4x+3)

TEST FOR FACTORS

To determine whether ax²+bx+c can be factorise into two linear factors evaluate the expression (D = b²-4ac)

If b²-4ac>0 {can be factorised into two linear factors. It has two real root}

If b²– 4ac<0{cannot be easily factorised into linear factor}

If b²– 4ac = 0 {it can be factorised into linear factor}

Example

Test the expression for factors

2x² +7x + 3

a = 2, b = 7, c = 3

b² – 4ac = D

7² – 4(2)(3) = D

D = 49 – 24 > 0

D>0

Therefore 2x² + 7x + 3 can be factorised into linear factor (2x+1) (x+3)

SOLUTION OF QUADRATIC EQUATION

There are several methods for solving a quadratic equation

1. Factorisation

2. Complete the square method

3. Formula method

1. Factorisation method

Example

Solve the equation: x² – 3x + 2 = 0

Solution

x² – 3x + 2 = (x – 1) (x – 2)

The equation may be written as

(x – 1) (x – 2) = 0

We wish to find a value such that when it is substituted for x in the member on the left, the product is zero, such may be found by solving either of

x –1 = 0 or x –2 = 0

x = 1 or x = 2

Example

Solve the equation 2x² + 5x – 12 = 0

Solution

2x² + 5x – 12 = 0

2x² + 8x – 3x – 12 = 0

2x(x+4) -3(x+4) = 0

(2x–3)(x+4) = 0

2x – 3 = 0 or x + 4 = 0

x = or x = –4

2 By completing the square method

Example: Solve x²+ 11x + 18 = 0

Solution

x² + 11x + 18 = 0

Subtract 18 from both side

x² + 11x = –18

Add to each side the square of half the coefficient of x

Example

Solve 4x² + 6x – 1 = 0 using completing the square method.

Solution

4x² + 6x – 1 = 0

Divide through by 4

Add ¼ to both sides

Add to each side the square of half the coefficient of x

Take the square root of both sides

Example

Solve ax²+bx+c = 0 using completing the square method

Solution

ax²+bx+c = 0

Subtract c from both sides

ax² + bx = − c

Divide through by a

Add to each side the square of half the coefficient of x

take the square root of both sides

This is called quadratic formula

3. By formula Quadratic

Example

Solve: 5x² + 12x + 3

a = 5, b = 12, c = 3

THE RELATION BETWEEN THE ROOTS OF A QUADRATIC EQUATION AND ITS COEFFICIENT

Sum and Product of Root

Consider the quadratic equation ax²+bx+c = 0,Where a, b, and c are constant such that a o,

Then

let represent these distinct root by and

Let b² – 4ac = D

Sum of roots

Product of root

Substitute D = b² – 4ac

Hence if ax²+bx+c = 0, a, b, and c are constants and a 0, then

The general quadratic equation ax²+bx+c = 0

Can be rewritten as

If the roots of the equation are and then the above can be written

(x – ) (x – ) = 0

x² – ( + ) x + = 0 - - - - - (2)

By comparing coefficient of like terms(i) and (ii)

( + ) =

+ =

and =

Example

Find the sum and product of the roots for each of the following

1. 5x² – 25x + 26 = 0

2. 3x² + 4x + 6 = 0

Solution

1. 5x² – 25x + 26 = 0

a = 5, b = −25, c = 26

Let and be root of the equation

Sum of roots: + =

Product of root: =

2. 3x² + 4x + 6 = 0

a = 3, b = 4, c = 6

Let and be the roots of the equation

Sum of roots: + =

Product of roots: =

3. x² + 6x – 8 = 0

a = , b = 6, c = – 8

Let and be the roots of the equation

Sum of roots:

Product of roots:

Example

Find the quadratic equation whose roots are

(a) 5 and –6

(b)

Solution

(a) The quadratic equation whose roots

and is x²–(sum of roots)x + product of roots = 0

Sum of root: + = 5 + (−6) = –1

Product: = 5 × (−6) = –30

The quadratic equation whose root is 5 and -6

is x² – (−1) x + (−30) = 0

x² + x – 30 = 0

(b)

Sum of roots .

Product of roots:

The quadratic equation becomes

6x² – 13x + 6 = 0

SYMMETRIC PROPERTIES OF ROOT

Consider the quadratic equation ax²+bx+c=0, a 0

Let , be roots of the equation

Sum of roots: + =

Product of roots: =

Certain relation involving and can

be determined from + and even when we do not know and distinctly, such relations are

said to be symmetric

e.g

whereas are not symmetric

NOTE: The roots and are symmetrical if when interchanged , the relationship remain the same or is multiplied by – 1

Example

If , determine whether or not each of the following is symmetric

Solution

Example

If are roots of the equation 2x²–7x –3=0. Find the value of

(a)

(b)

(c)

(d)

(e)

(f)

(g)

Solution

2x² – 7x – 3 = 0

a = 2, b = –7, c = – 3

IMPORTANT EXPRESSION

FORMATION OF NEW EQUATION

At times, it is required to from new equation from existing roots.

Let the existing roots be and , whose equation is

x² – x + , the new equation , let the roots be and

Example

If and are roots of the equation 2x² –7x –3 = 0 Obtain the equation that roots of which are

Solution

2x² – 7x – 3 = 0

a = 2, b = 7, c = 3

Sum of roots:

Product of roots

Product of root ,

The equation will be

Example

Given that the roots of the equation 3x² – 11x + 5 = 0 are and . Form the equation whose roots are

Solution

3x² – 11x + 5 = 0

(a = 3, b = –11, c = 5)

Let the roots be and

We seek an equation whose roots are

Example

The roots of the quadratic equation are , while those of the quadratic equation

2x² – 8x + 4 = 0 are Find the equation whose roots are

Let the equation be

ax² + bx + c =

Sum of roots

Example

If the roots of the equation x²+ bx + c = 0 are ,

and the roots of the equation

Show that the equation whose roots are

Solution

Example

From the quadratic equation for which the sum of the roots is 5 and the sum of the squares of the roots is 53

Solution

Example

If one of the roots of the equation ax²+ bx + c = 0 is four times the other show that 4b² – 25ac = 0

Solution

Example

One root of the equation x²-px + q = 0 is the square of the other. Show that P³ – q(3p+1) – q² = 0 provided q 1

Example

If one root of the equation x² – px + 8 = 0 is two more than the other. Find the value of p

Example

The roots of the equation x² +2ax + b = 0 differs by 2. Show that a² = 1+b

Example

Let and be the roots of the equation

5 + 12x + 4x² = 0

is defined by the function S_2n+1= where n is a natural number

Show that 78S₇ + 31S₉ = 0 and hence form an equation whose roots are S₃ and S₅

Solution

Sum of new roots: S₃ + S₅ = 2 − 6 = −4

Product of new roots: S₃. S₅ = 2× (−6) =−12

New equation formed

x² – (sum of roots)x + product of roots = 0

x² + 4x −12 = 0

NATURE OF ROOTS

Nature of roots has to deal with the type of roots the equation has, i.e. whether they are real or imaginary, rational or irrational, equal or unequal consider the solution of

Three constraints can be placed on the value of D

(1) D > 0

(2) D < 0

(3) D = 0

Note D = b² – 4ac is called the Discriminant of the equation and it is used to determine the nature of roots of a quadratic equation.

1. If b²- 4ac > 0: We shall obtain two real and distinct roots of the equation

The graph of ax² + bx + c = 0 crosses the x – axis at two distinct points

a>0

2. If b²–4ac<0, we shall obtain no real root or the equation has complex roots or imaginary roots

No real root

3. If b² – 4ac = 0: The roots are real and equal.

The equation can be said to form a perfect square

The roots are equal to

Example

Examining the diseriminant, describe the nature of roots of the following

1. 2x² – 7x + 3 = 0

2. x² – 6x + 9 = 0

3. 2x² – 3x + 2 = 0

Solution

1. 2x² – 7x + 3 = 0

a = 2, b = -7, c = 3

D = b² – 4ac

D = (-7)² – 4 × 2 × 3 = 49 – 24

D = 25 > 0

Hence the roots of the equation are real and distinct

2. x² – 6x + 9 = 0

a = 1, b = - 6, c = 9

D = b² – 4ac

D = (-6)² - 4×1×9 = 36 – 36

D = 0

It has two equal roots. It is a perfect square

3. 2x² + 3x +2

a = 2, b = 3, c = 2

D = b² – 4ac

D = 3² – 4×2×2 = 9 – 16

= – 7 > 0

Hence the roots are not real but imaginary or complex

Example

For what value(s) of r does the equation

x² – (2+r)x + 25 = 0 have equal roots

Solution

For equal roots to exist b² – 4ac = 0

a = 1, b = – (2+r), c = 25

[– (2+r)]² – 4×1×25 = 0

4+4r+r² – 100 = 0

r² + 4r – 96 = 0

(r+12) (r-8) = 0

r = -12 or r = 8

Example

Find the value(s) k, if the equation

(k+3)x² – (11k+1)x+k = 2(k – 5) has equal roots

Solution

(k+3)x² –(11k+1)x + k – 2 (k-5) = 0

a = (k+3), b = – (11k + 1), c = k – 2 (k – 5)

For equal roots to exist b² – 4ac = 0

[– (11k+1)]² – 4 (k+3) [k – 2 (k – 5)] = 0

(121k² +22k + 1) – 4 (k + 3) (10 – k) = 0

121k² + 22k +1 + 4k² – 28k – 120 = 0

125k² – 6k – 119 = 0

(125k+119) (k-1)

k = 1 or k =

Example

Show that the roots of the equation

x² – 2 (h – 2) x + 2h – 10 = 0

are real if h be real

Find the possible values of h, when the roots of equation differ by 6

Solution

i. For the roots to be real, then b² – 4ac > 0

a = 1, b = 2 (h-2), c = 2h – 10

[2(h-2)]² – 4 (1) (2h-10) > 0

4(h² – 4h + 4) – 4 (2h – 10) > 0

h² – 4h + 4 – 2h + 10 > 0

h² – 6h + 14 > 0

(h – 3)² + 5 > 0 (Using complete the square method)

i.e, if h be real, the square of real quantity is always positive

Adding (1) and (3) together

2 = 6 + 2 (h – 2)

2 = 2h + 2

= h + 1

Subtract (3) from (1)

2 =2 (h – 2) – 6

2 = 2h –10

= h – 5

Substituting = h – 2 and = h – 5 into (2)

= 2h – 10

(h +1) (h – 5) = 2h – 10

(h + 1) (h – 5) = 2 (h – 5)

(h + 1) [(h+1) -2] = 0

(h – 5) (h –1) = 0

h = 5 or h = 1

Example

If a and b are real, prove that the roots of the equation

(3a – b)x² + (b – a) x – 2a = 0 are real

Solution

(3a – b)x² + (b – a)x – 2a= 0

For real root to exist

(b – a)² – 4 (3a – b) (–2a) > 0

b² – 2ab+a² + 24a² – 8ab > 0

b² – 10ab + 25a² > 0

(b -5a)²> 0

If a and b are real, the square of any real number is always positive

Example

Prove that if a, b and c are real roots of the equation (a²+b²)x² + 2(a²+b²+c²) x+(b²+c²) = 0 are also real

Solution

(a²+b²)x²+2(a²+b²+c²)x + (b²+c²) = 0

For real roots

[2(a²+b²+c²)]² – 4(a²+b²) (b²+c²) > 0

4(a⁴+b⁴+c⁴+2a²b² + 2a²c²+2b²c²)– 4[b⁴+ a²b²+a²c²+b²c²] > 0

a⁴+b⁴+c⁴+2a²b²+2a²c²+2b²c²– b⁴– a²b²– a²c²– b²c²> 0

a⁴+c⁴+a²b²+a²c²+b²c²> 0

If a, b and c are real, the square of any real number is always positive

Example

For what values of the does the equation

10x²+4x+1 = 2kx (2– x) have real roots

Solution

10x² + 4x + 1 = 4kx – 2kx²

10x²+2kx²+4x – 4kx+1=0

(10+2k)x² + (4–4k)x +1 = 0

For real roots to exist

(4-4k)² – 4(10+2k) (1) > 0

16 – 32k +16k² – 40 – 8k > 0

16k² – 40k – 24 > 0

2k² – 5k – 3 > 0

(k-3) (2k+1) > 0

We draw our table to show the signs of the individual factor which change sign at and 3

	k <	<k < 3	k >3
k – 3	–	–	+
2k +1	–	+	+
(k-3)(2k+1)	+	–	+

So the value(s) of k to have real roots is k < or k > 3

Example

For what values of h does the equation x² – (4+h)x + 9 = 0 have real roots

Solution

x² – (4+h)x + 9 = 0

For real root to exist

[– (4+h)]² – 4 (1) (9) > 0

16 + 8h + h² – 36 > 0

h² + 8h - 20 > 0

(h + 10) (h – 2) > 0

We draw our table to show the sign of the individual factor which change sign at – 10 and 2

	h< - 10	-10 < h < 2	h > 2
(h + 10)	–	+	+
(h -2)	–	–	+
(h+10)(h – 2)	+	–	+

For real roots to exist h < – 10 or h > 2

QUADRATIC FUNCTION

An algebraic expression of the form ax²+bx+c is called quadratic function of x, where a 0

The graph of such function is a parabola

a>0

a<0

Minimum Maximum

If a > 0, e.g y = 5x² + 2x – 7, the function has a minimum

at the bottom of the curve.

If a < 0 , e.g y = –3x² – 4x +2, the function has a maximum value at the top of the curve

SIGN OF QUADRATIC FUNCTION

Given a quadratic function ax²+bx+c, where a, b, c are real and a 0 it important to know the condition when the quadratic expression is positive or negative, if b² – 4ac > 0

Let be the roots of the equation ax² + bx + c = 0. Then ax² + bx + c = a(x – )(x – ) where is taken as being greater than

We shall consider two cases

CASE 1: In this case the roots are real and unequal > when x > , (x – ) is positive and (x – ) is positive a(x – )(x –- ) ax²+bx+c is positive

when x < , (x – ) is negative and (x – ) is negative

a(x – )(x – ) ax²+bx+c is positive

The product of (x – )(x – ) will always be positive.

ax²+bx+c a(x –- )(x – ) positive

Hence the expression has the same sign as a

when > x > , (x – ) is negative and (x – ) is positive

a(x – )(x – ) ax²+bx+c is negative

Thus if a is positive, ax²+bx+c is positive for x >

or < and negative > >

CASE 2: a < 0 (a is negative)

As in case (1) , if a be negative ax²+bx+c is positive for

> x >

when > x > , (x – ) is negative and (x – ) is positive

a(x - )(x - ) ax²+bx+c is positive and negative

if x > or x <

We can understand this better using a tabular form

If a is positive

	x > a		x <
Sign of a	+	+	+
Sign of	+	–	–
Sign of	+	+	–
Sign of ax²+bx+c	+	–	+

If a is negative

	x > a		x <
Sign of a	–	–	–-
Sign of	–	–	–
Sign of	–	+	–-
Sign of ax²+bx+c	–	+	–

Example

Determine the sign of the function 16x² – 24x +10

Solution

As x is real , the term is always positive and hence the function is positive for all real values of x

Example

Show that 6x² – 5x +1 >0 for all values of x

Solution

As x is real , the term is always positive, and hence the function is positive for all real values of x

Example

Show that 6x²-5x+1>0 for all values of x

Solution

As x is real, the term is always positive, always be greater than and hence the function is greater than 0 for all real values of x.

MAXIMUM AND MINIMIUM VALUES

Given the quadratic expression ax²+ bx + c where a,b,c are real and a 0

We can notice that is a constant and is positive for all real x. Thus the nature of the expression depends partly on the value of x

We shall consider two cases

Case (i) when a > 0 (a is positive)

The expression ax²+bx+c has the least value when x = . In this case the value of quadratic expression will be a minimum and this minimum value is

Case(ii) when a<0 (a is negative)

The maximum value occurs when .

In this case the value of quadratic expression will be a maximum and this maximum value is

Example

i. if x be real, find the maximum value of 5+8x -4x²

ii. Find the least value of 4x²-5x+2 and hence deduce the greatest value of

Solution

5 + 8x – 4x² 5 + 4 (2x – x²)

5 – 4 (x² – 2x)

5 – 4 (x² – 2x +1) +4

9 – 4 ( x – 1)²

Since x is real, (x-1)² is always positive the expression has its maximum when x = 1

The maximum value of 5+8x-4x² is 9

Since is always positive, we see that

4x² – 5x + 7 has the least value when

The expression has its greatest value when 4x²-5x+2 has the least value of . Thus the greatest value of is = 16

QUADRATIC FRACTION

A quadratic fraction is of the form where f(x) and g(x) is either quadratic in x or linear in x, one at least being quadratic expression (or function)

e.g

We can find the limit between which must lie. We well illustrate this with the following examples

Example

Show that if x is real cannot lie between 1 and 4

Solution

x²-6x+5 = k (x²-6x+8)

x²-6x+5 = kx²-6kx+8k

x²(1-k) – 6 (1-k)x + (5-8k) = 0

Since x is real

[-6(1-k)]² – 4(1-k)(5-8k) > 0

36(1-2k+k²) -4 (5-13k+8k²) > 0

36k² – 72k +36 – 20 + 52K – 32k²> 0

4k² – 20k + 16 > 0

k² – 5k +4 > 0

(k – 1)(k – 4) > 0

If k lies between 1 and 4, the L.H.S of the above inequality is negative and for all other value of k the L.H.S is positive. hence, for real value of x, the expression cannot lie between 1 and 4

Example

Find the minimum value of for real values of x

Solution

6x² – 22x + 21 = 5kx² – 18kx + 17k

(6 – 5k)x² – 2 (11 – 9k) + (21 – 17k) = 0

Since x is real

[–2(11–9k)]² – 4 (6–5k)(21–17k) > 0

4(81k² – 198k+121) – 4(85k² – 207k + 126) > 0

4(81k² – 198k + 121 – 85k² + 207k – 126) > 0

4(–4k² + 9k – 5) > 0

–4k² + 9k – 5 > 0

4k² – 9k + 5 < 0

(4k – 5) (k – 1) < 0

The k lies between 1 and the L.H.S of the above inequality is negative and for all other values of x the L.H.S is positive. Hence for real values of x, the expression must lie between 1 and

Hence the minimum value of k is 1