Intermediate Quadratic Equations

[Dragonballful23]

Find the solution set for these:






This one is the hardest....

[MAS1]

Originally Posted by Dragonballful23

Find the solution set for these:






This one is the hardest....

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

3x^2 + 4x + 1 = 0
(3x + 1)(x + 1) = 0
3x +1 = 0
3x = -1
x = -1/3
x + 1 = 0
x = -1
So x = -1/3, -1

x^2 + 3x = -2
x^2 + 3x +2 = 0
(x + 1)(x + 2) = 0
x = -1, -2

2x^2 + 10x = -15
x^2 + 5x = -15/2
x^2 + 5x + 25/4 = -15/2 + 25/4
(x + 5/2)^2 = -5/4
sqrt((x + 5/2)^2) = sqrt(-5/4)
x + 5/2 = + or - i(sqrt(5)/2)
x = -5/2 + or - i(sqrt(5)/2)

[john123]

To find roots of any quadratic equation ,
one formula is there
if ax^2+bx+c is the given quadratic equation then roots are given by
x= ( -b±√b^2-4ac)/2a

[Dragonballful23]

Originally Posted by john123

To find roots of any quadratic equation ,
one formula is there
if ax^2+bx+c is the given quadratic equation then roots are given by
x= ( -b±√b^2-4ac)/2a

There is also a shortcut(my teachers tell me not use ) -b/2 plus/minus square root of (b/2)^2-ac



Hey Mr.Hui can you tell me where the sqared option is?