Can someone explain how to solve this equation: 4x^2 +2x + 4?
"solve" is vague. what is it you are actually trying to do?
I need to understand how to find the value of x.
Hello
Hi Dan. Hi Amistre. It is an honor to be helped by two of OpenStudy's best helpers.
I think your equation is most likely 4x^2 +2x + 4 =0 or 4x^2 +2x + 4 = y you have to find the x based on based on the given value for y
as is, there is no "equality" to it, so x is any value on its domain
Sorry, it is = 0. My bad
I am just starting to learn quadratics.
the simplest solution then is to apply the quadratic formula have you learned the formula?
But I want to also understand how to solve it with the other non formula way. Do you have a strategy? If you want to use another example that is fine too.
Completing the square is the main way, that is where the quadratic formula comes from
I just made this one up. But I have noticed that there are people in my class, and my teacher - that seem to know how to group things etc
the strategy is something called completeing the square, that im sure dan will explain to you
Please do.
4x^2 +2x + 4 =0 okay so completing the square means you are going to have to put it in this form (ax+b)^2=c
Well, this has no real root.
can you expand the expression on the left and see what happens (ax+b)^2=c
a^2x^ +b^2 + 2axb
4x^2 +2x + 4 =0 a^2x^ +2abx+b^2 = c okay look here we must first make sure we need the criteria for the coefficients of x, the b^2 and c we can compensate for as they are just constants and we can add constants to both sides of the equation
so we know that a^2 = 4 and 2ab = 2
this means a=2 and b=1/2
any questions so far?
That is pretty clever.
Thanks
Still with you. But I did rate you.
4x^2 +2x + 4 =0 we are going to rewrite the 4x^2+2x part now (2x+1/2)^2 +4 = +1/4 i am adding the b^2 = 1/4 to both sides, the other 1/4 is coming from the expansion of the squared bracket
(2x+1/2)^2 = +1/4 - 4 =-3.75 now we can solve for x 2x+1/2 = +- sqrt(-3.75) x=+-(sqrt(-3.75) - 1/2)/2 notice the + or - coming up at the squareroot point and also notice that negative value in the square root so your roots are imaginary in this case
okay thanks :) please click on that golden bell on top to rate me
if you understand this question
|dw:1430236845485:dw|
Join our real-time social learning platform and learn together with your friends!