quadratic equation

Quadratic Equation Simplified Notes

Consider the quadratic equation 2x2 + 3x - 2 = 0. ----( *** )

We can solve x by :-

( I ) Factorization

1 st Trial & Error

i.e 2x2 - 2 + 0 = 0 ≠ ( *** )

2nd Trial & Error

i.e 2x2 - 2 + 3x = 0. = ( *** )

( *2x - 1* ) ( **x + 2** ) = 0

x = 1/2 or x = -2

( II ) Completing the Square

2x2 + 3x - 2 = 0

x2 + 3/2 x - 1 = 0

x2 + 3/2 x = 1 ==============( * )

Consider the coefficient of x;

i.e coefficient of x = 3/2 =========( ii )

Divide both sides of equation ( ii ) by 2 :-

i.e

{coefficient of x} / 2= ( 3/2 ) / 2

= ( 3/4 ) ========( iii)

square, both side of equation ( iii ) :-

i.e [{coefficient of x} / 2 ]2 = ( 3/4 )2

ADD ( 3/4 )2 to both sides of ( * )

( 3/4 )2 + x2 + 3/2 x = 1 + ( 3/4 )2

OR

x2 + 3/2 x + ( 3/4 )2 = 1 + ( 3/4 )2

(x + 3/4 )2 = 1 + ( 9/16)

= 25/16 note: ( a + b )2 = a2 + b2 +2ab

(x + 3/4 )= + 5/4

x = 5/4 - 3/4 or x = -5/4 -3/4

x = 2/4 = 1/2 or x = -8/4 = -2

( III ) By using the Formula

x = [ -b + √ ( b2 -4ac ) ]/ 2a

= [ -3 + √ ( 32 -4( 2 )( -2 ) ]/ 2(2)

= [ -3 + √ ( 9 + 16 ]/ 4

= [ -3 + √ ( 25 ]/ 4

= [ -3 + 5 ]/ 4

= ( -3 + 5 ) / 4 or = ( -3-5 ) /4

= 2/4 = 1/2 or = -8 / 4 = -2

Hence, -2 and 1/2 are roots ( two different real roots) of

an equation 2x2 -3x - 2 = 0.

Note:

if b2-4ac > 0 ; the equation has two different real roots if b2-4ac = 0 ; the equation has two equal real roots if b2-4ac < 0 ; the equation does not have real roots

>> 2x2 -3x - 2 = 0 has two different real roots :-

>>> b2-4ac = -32 -4( 2 )( -2)

= 9 + 16

= 25 > 0

In general, if p & q are roots of a quadratic equation:-

then,

( x - p )( x - q ) = 0

x2 - px-qx + pq = 0

x2 - ( p + q )x + pq = 0

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

Question 1

( i ) A quadratic equation x2 + px + 9 = 2x has two equal roots

Find the possible values of p

Solution Method

the equation has two equal real roots :b2-4ac = 0

Solution

x2 + px + 9 = 2x

x2 + px- 2x + 9 = 0

x2 + (p - 2)x + 9 = 0

The above equation has two equal real roots :

b2-4ac = 0

( p-2)2 -4(1)(9) = 0

( p-2)2 - 36 = 0

( p-2)2 - = 36

p - 2 = +6

p-2 = 6 or p-2 = -6

p = 8 or p = -4

( ii ) Find the range value of k so that the

quadratic equation x2 + kx - 3 = -2k has no real roots

Solution Method

the equation has no real roots :b2-4ac = < 0

Solution

x2 + kx -3 = -2k

x2 + kx + 2k - 3 = 0

x2 + kx +( 2k - 3 ) = 0

The above equation has no real roots :

b2-4ac < 0

( k )2 -4(1)(2k-3) < 0

k2 -8k + 12 < 0

( k -2 )( k-6) < 0

k < 2 ; k < 6

2< k < 6

( iii )Determine the set of values of m so that the

equation x2 + ( m + 4 )x + ( m2 + m + 3 ) = 0

has real roots

Solution

For real roots, b2-4ac = 0

( m+ 4 )2-4(1)(m2 + m + 3 ) = 0

m2 + 8m + 16-4m2 -4m -12 = 0

-3m2 + 4m + 4 = 0

3m2 - 4m - 4 = 0

m = [ -b + √ ( b2 -4ac ) ]/ 2a

= [ 4 + √ ( 16 + 48 ) ]/ 6

= [ 4 + √ ( 64 ) ]/ 6

= [ 4 + 8 ]/ 6

m = (4 -8 )/6 or m = (4+8)/6

m = -2/3 or m = 2

i.e -2/3 < m < 2

( iv ) The quadratic equation hx2 + kx +9 = 0 , where h and k are

constants, has two equal roots.

Express h in terms of k

Solution

For equal roots, b2-4ac = 0

k2-4(h)(9) = 0

k2 = 36h

h = k2 / 36

( v ) Given that 3/2 and - 15 are the roots for the given quadratic

equation.Write down that quadratic equation in the

form of hx2 + kx +9 = 0

Solution

( x -3/2 ) ( x -(-15) ) = 0

( x-3/2)(x + 15 ) = 0

x2 + 15x - ( 3/2)x - 45/2 = 0

x2 + (30/ 2 )x - ( 3/2)x - 45/2 = 0

2x2 + 17x - 45 = 0

Assessment

( i ) The quadratic equation, px2 + qx + 3 = 0,

where p and q are constants, has two equal roots.

Express p in terms of q

Ans : p = q( 12)1/2

( ii ) The quadratic equation x2 + x = 2px -p2 ,

where p is a constant, has two different roots.

Find the range values of p

Ans : p < 1/4

( iii ) The quadratic equation x2 + px + 9 = 2x ,

has two equal roots.

Find the possible values of p

Ans : p = 8 or p = -4

( iv ) The roots of quadratic equation 2x2

+ px = -q ,

are -5 and 4

( a ) Find the values of p and q

( b ) the range of values of k such that

2x2

+ px = k -q does not have any real roots

Ans : ( a ) p = 2, q = -40

( b ) k < 81/2

Note :If m and n are the roots for a quadratic equation

ax2 + bx + c = 0 ====( 1 )

a # 0

OR ax2 - ( - bx ) + c = 0

Divide both side of ( * ) by a:-

x2 - ( -b/a)x + c/a = 0 ----------( 2 )

Then,

sum of root = ( m + n )

= -b/a

Product of the roots = mn

= c/a

Conclusion

If m and n are the roots of a quadratic equation

ax2 + bx + c = 0 ====( 1 )

Change ( 1 ) into ( 2 ) :

x2 - ( -b/a)x + c/a = 0 ----------( 2 )

So that, sum of root =( m + n )

= -b/a


= c/a

Question 2

( i ) If a and a/2 are roots of a quadratic equation

x2 + px + q = 0

Show that 2p2 = 9q

Solution Method ( and also Solution )

Let m & n are the roots of the equation

x2 + px + q = 0

x2 - ( - p )x + q = 0 ===( 1 )

Sum of the roots = ( m )+ ( n )

= -p ====( 2 )


= q ==( 3 )

But, ( a ) & ( a/2 ) are also roots of ( 1 )

Hence, sum of root = ( a )+ ( a/2 )

= 3a/2 === ( 4 )

Product of root = ( a )( a/2)

= a2 /2 ==== ( 5 )

From ( 2 ) & ( 4 ) :

-p = 3a/2

a = -2p/3 ===== ( 6 )

From ( 3 ) & ( 5 ) :

q = a2 /2 ======( 7 )

( 6 ) in ( 7 ) : q = ( -2p/3 )2 / 2

= 2p2 / 9

9q = 2p2 ( ii ) GIven that p and q are two roots of the

equation 2x2 = - ( 3x + 4 ) :-

a) Find 1/p2 + 1/q2

b) Show that 4p4 = -(16 + 7p2 )

Solution Method ( a ) ( and also Solution )

p and q are the roots of the equation

2x2 = - ( 3x + 4 )

2x2 + 3x + 4 = 0

2x2 - ( - 3 )x + 4 = 0

x2 - ( - 3/2 )x + 2 = 0 === ( 1 )

Sum of roots = p + q

= -3/2 ===== ( 2 )

Product of roots = pq

= 2 ====== ( 3 )

Then,

1/p2 + 1/q2 = (q2 + p2) / (p

2q2

)

= *( p + q )2

- 2pq / (pq )2 ===( 4 )

Note : *(p + q)2

= (p2 + p2)

+ 2pq

Subsitute ( 2 ) & ( 3 ) in ( 4 ) :-

1/p2 + 1/q

2 = [( p + q )

2 - 2pq ]/ (pq )

2

= [( -3/2 )2

- 2( 2 )]/ (2 )2

= - 7/16

Solution ( b )

From ( a ) 1/p2 + 1/q

2 =

- 7/16

1/p2 =

- 7/16 - 1/q2

= - 7/16 - 1/( 2/p )

2

= - 7/16 - (p

2 / 4)

= ( - 7 -4p

2 )/ 16

16 = -7p 2 -4p

4

4p 4

= -7p 2 -16

= - ( 7p 2 + 16 )

Assessment

( i) Given that p/3 and q/3 are two roots of the

equation 6x2

= 3x + 2 :-

a) Find 1/p + 1/q

b) Show that 2q2

= 6 + 2q

Ans:

( i) Given that p and q are two roots of the

equation 2x2

= 3x -4 :-

a) Form an equation whose roots are

p- q and q-p

b) Show that 4p3

= p- 12

Ans : ( a) 4x2 - 23 = 0

Question 3

Given that the quadratic function 2x2 - px + p + 1 = 0

has roots of m and n

If 4+ ( m2 +n 2 ) = 9

Find the positive value of p.


m and n are the roots of

2x2 - px + p + 1 = 0 --( * )

x2 - ( p/2 )x + 1/2 ( p + 1 ) = 0

Sum of the roots = m + n

= p/2 ==== ( 1 )


= 1/2( p + 1 ) ====( 2 )

Given that :-

4+ ( m2 +n 2 ) = 9 ;note : ( m +n )2 = m2 + n2 + 2mn

4+ ( m +n )2 - 2mn = 9

( m +n )2 - 2mn - 5 =0 ==== ( 3 )

Substitute ( 1 ) & ( 2 ) in ( 3 ) :-

( p/2 )2 - 2 [ ( 1/2(p + 1 )]-5 =0

p2 / 4 - p - 1 -5 =0

p2 / 4 - p - 6 =0

p2 - 4p - 24 =0

p = [ -b + √ ( b2 -4ac ) ]/ 2a

= [ - (- 4 ) + √ ( -42 -4( 1 ) ( - 24) ]/ 2( 1 )

= [ 4 + √ ( 16 + 96 ]/ 2

= [ 4 + √ ( 112)]/ 2

=[ 4 + ( 10.58 )]/ 2

Since p > 0

Thus p = [ 4 + 10.58 ]/ 2

= 7.03

Question 4

Given that p and q are the roots of the quadratic equation

2x2 -8x + 3 = 0

Form a quadratic equation with roots p + 1/p and q + 1/q


p and q are the roots of

2x2 - 8x + 3 = 0 --( * )

x2 - ( 4 )x + 3/2 = 0

Sum of the roots = p + q

= 4 ===== ( 1 )

Product of the roots = pq

= 3/2 ==== ( 2 )

Let m = ( p + 1/p ) and n = ( q + 1/q ) be the roots of the

equadratic equation

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

where

Sum of the roots = m + n

= ( p + 1/p ) + ( q + 1/q ) === ( 3 )


= ( p + 1/p ) ( q + 1/q ) ====== ( 4 )

From ( 3 ) :-

Sum of the roots = ( p + 1/p ) + ( q + 1/q )

= 1/p( p2 + 1 ) + 1/q( q2 + 1 )

= 1/pq [ q( p2 + 1 ) + p( q2 + 1 ) ]

= 1/pq [ ( qp2 + q ) + ( pq2 + p ) ]

= 1/pq [ ( qp2 + pq2 ) + ( p + q ) ]

= 1/pq [ pq( p + q ) + ( p + q ) ]

= 1/pq [ pq( p + q ) + ( p + q ) ]

= 1/pq[ (p + q )(pq + 1 )]

=2/3[ (4 )( 3/2 + 1 )]

= 2/3[ (4 )( 3/2 + 1 )]

=20/3 ===== ( 5 ) From ( 4 ):-

Product of the roots =( p + 1/p ) ( q + 1/q )

=1/p ( p2 + 1 )( q2 + 1 )1/q

=1/pq ( p2 + 1 )( q2 + 1 )

=1/pq [ 2( pq )2 + 1 ]

= 2/3 [ 2( 3/2 )2 + 1 ]

= 2/3 [ 2( 3/2 )2 + 1 ] = 26/9 ===== ( 6 )

Substitute ( 5 ) & ( 6 ) in ( * ) becomes :-

x2 - 20/3x + 26/9 = 0

9x2- 60x + 26 = 0

Asessment

( i ) Given that p and q are the roots of the quadratic

equation 3x2

+2x - 1 = 0

Form a quadratic equation with

roots 2p + 3 and 2q + 3

Ans: 3x2

-14x + 11 = 0

( ii ) Given that p and q are the roots of the quadratic

equation 2x2

-3x - 6 = 0

Form a quadratic equation with

roots p/3 and q/3

Ans : 6x2

-3x - 2 = 0

Question 5

( i ) Given that one of the roots of the equation

x2 + px + 12 = 0 is three times the other root,

find the possible value of p

Solution Method ( i ) and also Solution ( i )

Let r and s be the roots of

x2 + px + 12 = 0

x2 - ( -p)x + 12 = 0 ===== ( * )

Sum of the roots = r + s

= -p ===== ( 1 )

Product of the roots = rs

= 12 ==== ( 2 )

( Given that one of the root = 3times the other

i.e either s = 3r or r =3s )

Let s = 3r

Then ( r ) & s ( = 3r ) are also roots of equation ( * )

Sum of roots = r + s

= r + 3r

= 4r ==========( 3 )

Product of the roots = r ( s )

= r( 3r)

= 3r2 ======( 4 )

From ( 1 ) & ( 3 ) :-

4r = -p

p = -4r ===== ( 5 )

From ( 2 ) & ( 4 ) :-

3r2 = 12

r2 = 4

r = + 2 ==== ( 6 )

Substitute ( 5 ) in ( 6 ) :-

When r = 2, p = -4( 2 ) = -8

r = -2, p = -4( -2) = 8

( i i ) Given that 4/p and 4/q are the roots

of the quadratic equation

kx ( x - 2 ) = - ( 2m + 7x )

If p + q = -3/2 and pq = 4.

Find the values of k and m

Solution Method ( ii )and also Solution ( ii )

Let p & q are roots of the quadratic equation

kx ( x - 2 ) = - ( 2m + 7x )

kx2 - 2kx = - 2m- 7x

kx2 - 2kx + 7x + 2m = 0

kx2 - (2k - 7 )x + 2m = 0

x2 - [ (2k - 7 )/k ] x + 2m/k = 0 ====== (* )

where

Sum of the roots, = p + q

= (2k - 7 )/k ====== ( 1 )

Product of the roots = pg

= 2m/k ==========( 2 )

Since ( 4/p ) & ( 4/q ) are also roots of ( * )

Thus,

Sum of the roots =( 4/p )+ ( 4/ q )

= 1/ pq [ 4q + 4p ]

= 4/pq [ ( p + q ) ]

= 4/4 [ -3/2 ]

= -3/2 ======= ( 3 )

Product of the roots = ( 4/p ) ( 4/q )

= 16/pq

= 16/4

= 4 ========= ( 4 )

From ( 1 ) and ( 3 );

(2k - 7 )/k = -3/2

2(2k -7) = -3k

7k -14 = 0

k = 14/7 = 2

From ( 2 ) and ( 4 );

2m/k = 4

2m/ 2 = 4

m = 4

Assessment

( i ) Given that p/3 and q/3 are the roots of the quadratic equation

kx ( x - 1 ) = x + m

If p + q = 4 and pq = 3.Find the values of k and m

Ans : k = 3, m = -1

( i i) Given that 3 and p are the roots of the quadratic

equation (x + 1 ) ( 2x + 4 ) = q(x - 1 ) ,

q is a constant .Find the values of p and q

Ans : p = 4,q = 20

Question 6

Find the value of p so that the quadratic equation

( 3 - p )x2 -2( p + 1)x + p+1 = 0 , has two equal roots.

Hence, find the roots based on p that you have obtained

Solution

( 3 - p )x2 -2( p + 1)x + p+1 = 0

For equal roots, b2-4ac = 0

[-2( p + 1 )]2 -4( 3-p )(p + 1 )= 0

4( p + 1 )2 -4( 3p + 3 - p2 -p)= 0

( p + 1 )2 -( 3p + 3 - p2 -p)= 0

p2 + 2p + 1 - (-p2 +2p + 3 ) = 0

p2 + 2p + 1 + p2 -2p - 3 ) = 0

2p2 - 2 = 0

p2 - 1 = 0

p2 = 1

p = +1

When p = 1, The quadratic equation is :-

( 3 - p )x2 -2( p + 1)x + p+1 = 0

( 3 - 1 )x2 -2( 1 + 1)x + 1+1 = 0

2x2 -4x + 2 = 0

( x - 1 )2 = 0

x = 1

When p = -1, The quadratic equation is :-

( 3 - p )x2 -2( p + 1)x + p+1 = 0

( 3 + 1 )x2 -2( -1 + 1)x - 1+1 = 0

4x2 = 0

x2 = 0

x = 0

Question 7

The quadratic function f(x) = 2x2+ 5x - 3 can be expressed in the

form f(x) =(x + m )2 -n, where m and n are constants

Find the value of m and n

---------------------------------------------------------------------------

Solution Methods

f(x) =(x + m )2 -n, =====> Completing the square

Completing the Square

2x2 + 5x - 3 = 0

x2 + 5/2 x - 3/2 = 0

x2 + 5/2 x = 3/2 ================ ( * )

Consider the coefficient of x;

i.e coefficient of x = 5/2 ============== ( 1 )

Divide ,equation ( 1 ), both sides by 2 :-

{coefficient of x} / 2 = ( 5/2 ) / 2

= ( 5/4 ) ========== ( 2 )

Square ,equation ( 2 ) both sides :-

Thus [{coefficient of x} / 2 ]2 = ( 5/4 )2 ====( 3 )

ADD ( 5/4 )2 to both sides of ( * )

x2 + 5/2 x + ( 5/4 )2 = 3/2 + ( 5/4 )2 = 69/16

( x + 5/4 )2 = 69/16 ; note : ( a + b )2 = a2 + b2 + 2ab

( x + 5/4 )2 - 69/16 = 0

Solution

f(x) = 2x2+ 5x - 3

= x2 + 5/2 x -3/2

= x2 + 5/2 x -3/2 + { ( 5/4 )2 - ( 5/4 )2 }

= [x2 + 5/2 x + ( 5/4 )2 ] -3/2 - ( 5/4 )2

= ( x + 5/4 )2 -3/2 - ( 5/4 )2

= ( x + 5/4 )2 -69/16

comparing with

f(x) =(x + m )2 -n

m = 5/4 & n = -69/16

Question 8

The diagram shows the graph of a quadratic function y = f( x ).The straight line y = -9 is a tangent to the curve y = f ( x )

( a ) Write the equation of the axis of symmetry y of the curve

( b ) Express f ( x ) in the form of f(x) =(x + b )2 + c ,

where b and c are constants

Solution

Axis of symmetry , x = ( -1 + 5 )/2 = 2

i.e x = 2

f( x ) = ( x + b )2 + c

= ( x - 2 )2 -9

quadratic equation

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