Question

Will Give Medal!

Decide which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring.

−b b^2 − 4ac 2a

Use the part of the quadratic formula that you chose above and find its value, given the following quadratic equation:

4x^2 + 6x + 2 = 0

Answer 1

if b^2-4ac is a perfect square then the quadratic equation can be factored.

Answer 2

Ok... I dont completly understnad... how did you get that and also what do you do next..? @surjithayer

Answer 3

Oh i see where you got b^2-4ac now...

Answer 4

\[x=\frac{ -6 \pm \sqrt{6^2-4*4*2} }{ 2*4 }=\frac{ -6 \pm \sqrt{36-32} }{ 8 }=\frac{ -6 \pm2 }{ 8 }\]
find values of x

Answer 5

Oh ok so you solved the quadratic formula that they gave you... so now what do i do...

Answer 6

In the equation a=4 b=6 and c=2 so do i just plug that into either −b b^2 − 4ac 2a...?

Answer 7

Or do i plug it into all of them...

Answer 8

\[ax^2+bx+c=0,gives ~a(x-\alpha)(x-\beta)=0\]
where alpha and beta are zeros of ax^2+bx+c=0

Answer 9

ok...

Answer 10

So now i plug in my a b and c values into that equation??

Answer 11

a=4 and find zeros of quadratic or find values of x

Answer 12

can you show me the values of x

Answer 13

ok so

-1 and -0.375

Answer 14

\[x=\frac{ -6+2 }{ 8 }=\frac{ -4 }{ 8 }=-\frac{ 1 }{ 2 }\]
\[or~x=\frac{ -6-2 }{ 8 }=-\frac{ 8 }{ 8 }=-1\]
so \[4x^2+6x+2=0~gives~4\left( x+\frac{ 1 }{ 2 } \right)\left( x+1 \right)=0\]
\[4\left( \frac{ 2x+1 }{ 2 } \right)\left( x+1 \right)=0\]
\[or~2\left( 2x+1 \right)\left( x+1 \right)=0\]

Answer 15

i have make factors.
But as desired in the problem b^2-4ac=6^2-4*4*2=36-32=4=2^2 or(-2)^2
so it is a perfect square.
hence factors can be made.