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Chapter ( 2 )
Simultaneous equations and Quadratics
2.1 Simultaneous Equations
Solving simultaneous equations is finding the points of intersection of
the two graphs given by these equations.
Substitution Method
Example : π¦ = π₯
π₯2 + π¦2 = 8
Substitute π¦ = π₯ ππ π₯ 2 + π¦ 2 = 8 .
π₯2 + π₯2 = 8
2π₯ 2 = 8
π₯2 = 4
π₯ = ±2
πβππ π₯ = 2 , π¦ = 2 πππ π€βππ π₯ = −2 , π¦ = −2 .
Points of intersection are ( 2, 2 ) πππ ( −2 , −2 ).
2.2 Maximum and Minimum Values of a Quadratic Function
General form of a quadratic function is y = ax 2 +bx+c , where a ≠ 0.
Graph of a quadratic function
The graph of a quadratic function is called a parabola .
If a is positive ( i.e, a > 0 ) , the parabola opens up.
If a is negative ( i.e, a < 0 ) , the parabola opens down.
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shape
shape
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Complete square form and Vertex
π
π¦ = ππ₯ 2 + π π₯ + π = π ( π₯ 2 + π π₯ ) + π
π
2
π
π
2
= π [ π₯ 2 + π π₯ + (2π) − (2π) ] + π
π
2
π2
π
2
π2
= π[( π₯ + 2π ) − 4π2 ] + c
= π ( π₯ + 2π ) −
π
2
= a ( π₯ + 2π ) −
4π
4ππ
+ 4π
( π 2 − 4ππ )
4π
π
= π( π₯ − β )2 + π , where h = − ππ and k = −
( ππ − πππ)
Vertex of a parabola [ Turning point , Stationary point ]
( 1 ) Vertex from the complete square form
π¦ = a( x − h )2 + k
The vertex of the parabola is ( β, π )
b
Vertex = ( h, k ) = ( − 2a , −
b2 − 4ac
4a
).
( 2 ) Factor form and Vertex
y = ππ₯ 2 + π x + c = a( x – p )( x – q )
x- coordinate of vertex = h =
Vertex = ( h , k ) = (
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π+π
π+π
2
2
π+π
, y- coordinate of vertex = k = f ( 2 )
π+π
, π ( 2 ))
ππ
.
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Maximum and and Minimum value of y = ππ± π + π x + c and Vertex
( 1 ) If a > 0 , y = ax 2 + b x + c has the minimum value at the vertex.
π
y = a( x − h )2 + k , a >
0
Axis of symmetry, x = h
Minimum value = π = π (− 2π )
Vertex ( h, k )
Minimum point
( 2 ) If a < 0 , y = ax 2 + b x + c has the maximum value at the vertex.
π
Maximum value = π = π (− 2π )
Maximum point
Axis of symmetry, x = h
Vertex ( h, k )
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y = a( x − h )2 + k , a < 0
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2.3 Graphs Of π = | π( π ) | πππππ π( π ) ππ π ππππ
πππππ ππππππππ.
π( π₯ ) = ax 2 + bx + c , a ≠ 0
y = | f( x ) | = | ax 2 + bx + c |
1. First draw the graph of π¦ = π( π₯ ) .
2. Reflect in the x-axis the part of the curve π¦ = ax 2 + bx + c that is
below the x-axis .
π¦
π¦ = |π(π₯)|
π₯
π
π¦ = π( π₯ )
π¦
π¦ = |π(π₯)|
π₯
π
π¦ = π( π₯ )
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2.4 Quadratic Inequality
ax 2 + bx + c < 0
ax 2 + bx + c ≤ 0
ax 2 + bx + c > 0
ax 2 + bx + c ≥ 0
The solution set of an inequation ( y > 0 or y < 0 ) is the set of all x
which satisfies this inequation.
b2 − 4ac > 0
y = ax 2 + bx + c = a ( x – p ) ( x – q ) and p < q .
y = a(x − p)(x − q) ,a>0
y = a(x − p)(x − q),a<0
+
p
+
+
p
q
−
q
x
−
π¦ > 0 → π₯ < π ππ π₯ > π
x
−
π¦>0 →π<π₯< π
π¦ < 0 → π₯ < π ππ π₯ > π
π¦<0 →π<π₯<π
2.5 Roots Of A quadratic Equation
The solutions of a quadratic equation ππ₯ 2 + ππ₯ + π = 0 are called
the roots of this quadratic equation.
πΈπ₯πππππ ; 2π₯ 2 − 3π₯ − 5 = 0
( 2π₯ − 5) ( π₯ + 1 ) = 0
2π₯ − 5 = 0 ππ π₯ + 1 = 0
5
π₯=2
ππ π₯ = −1
5
πβπ ππππ‘π ππ π‘βπ πππ£ππ πππ’ππ‘πππ πππ − 1 πππ 2 .
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Quadratic Formula of ax 2 + bx + c = 0 .
ax 2 + bx + c = 0
ax 2 + bx = − c
b
x
a
= −
c
a
b
b 2
x +( )
a
2a
= −
c
a
x2 +
x2 +
(x+
b 2
2a
+ ( )
b 2
4ac
b2
) = − 2 + 2
2a
4a
4a
b
2
x+
b
2a
b2 − 4ac
4a2
( x + 2a ) =
= ± √
x = −
x=
b
2a
b2 − 4ac
4a2
±
√ b2 − 4ac
2π
−b ± √ b2 − 4ac
2a
Discriminant of a Quadratic Function
y = ax 2 + b x + c
π·ππ πππππππππ‘ = b2 − 4ac .
( i ) If π 2 − 4ππ > 0 ,
ax 2 + b x + c = 0 has two distinct real roots and so
the graph cuts the x- axis twice.
y = ax 2 + bx + c , a < 0
y = ax 2 + bx + c , a > 0
b2 − 4ac > 0
b2 − 4ac > 0
x
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x
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( ii ) If π 2 − 4ππ = 0 ,
ax 2 + b x + c = 0 has two equal real roots ( only one root ) and so
the graph is tangent to the x- axis at only one point.
y = ax 2 + bx + c , a < 0
2
y = ax + bx + c , a > 0
b2 − 4ac < 0
b2 − 4ac = 0
x
x
( iii ) If π 2 − 4ππ < 0 ,
ax 2 + b x + c = 0 has no real root and so
the graph does cut the x- axis .
y = ax 2 + bx + c , a < 0
y = ax 2 + bx + c , a > 0
b2 − 4ac < 0
b2 − 4ac < 0
x
x
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2.6
Intersection Of A Line And A Conic Section Curve
Conic Sections are Parabolas , Circles , Ellipses and Hyperbolas .
After substituting the line equation in conic section equation
( π΄ π₯ 2 + π΅π₯π¦ + πΆπ¦ 2 + π·π₯ + πΈπ¦ + πΉ = 0 ) , we will have the quadratic equation
ππ₯ 2 + ππ₯ + π = 0 .
Thus we can determine whether or not the line intersect the curve
as following:
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Discriminant
π 2 − 4ππ > 0
Two points of
intersection
Discriminant
π 2 − 4ππ = 0
Exactly one point of
Intersection
i.e. the line is tangent
to the curve.
Discriminant
π 2 − 4ππ < 0
No point of
intersection
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A maths ( 0606 ) Quadratic and Modulus
2.
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