Quadratic solutions
1.
2x2
+ 8x + 1 = 2[x2 + 4x] + 1
= 2[(x + 2)2 –
4] +1
= 2(x + 2)2 –
7
2. 3x2 - 5x – 2 =
= 
3. 2 + 3y – y2 = 
max.
value is 
occurring when 
3.
6 – x
– 2x2 = 
To solve let = 0
(x
+ 
) = 
x = 
and x = -2
check 6 – x – 2x2 =
0
(3 - 2x)(2 + x) = 0
so x = and x = -2
5. 
to find the
minimum value of f(y) first find the max. value of the denominator.
4 – y – y2
= 
the max. value
of the denom. is
f(y)
= 
this
minimum of f(y) occurs when y = 
6. ax2 + bx + c = 0 a[ x2 + 
] + c = 0
a[(
x + 
)2 - 
] + c = 0
a(
x + )2 - 
+ c = 0
a(
x + )2 = 
a(
x + )2 = 
a(
x + )2 = 
(
x + )2 = 
(
x + ) = 
x
= -
x
= 
which is the
quadratic formula
7. Sum of these roots is 
and the
product of the roots is 