quadratic equation
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Latih Tubi Matematik Tambahan - Persamaan KuadratikTRANSCRIPT
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
Product of the roots = mn
= 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 )
Product of the roots = mn
= 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.
Solution Method ( and also Solution )
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 )
Product of the roots = mn
= 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
Solution Method ( and also Solution )
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 )
Product of the roots = mn
= ( 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