Find the solutions for x:
4
x
2
+
8
x
+
3
=
0
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
nice and easy solution (unique)
nice solution @Mathh Mathh, didnot know about that rule thanks for sharing.
Log in to reply
I specifically made a simple problem to introduce brilliant community to it here .
Log in to reply
thanks for the contribution, btw it was a nice problem .
Ha ha used a different solution , i used vieta formulas for sum and product so x 1 + x 2 = − b / a and x 1 ∗ x 2 = c / a by vieta so x 1 + x 2 = − 2 and product is x 1 ∗ x 2 = 3 / 4 and we see that − 1 . 5 , − . 5 is the solution we need.
lol using vieta on such a problem
Descartes' Rule of Signs' says it can't have only positive solutions.that means if there is two solutions one must have to be negative
There was only one choice with all numbers negative.so,either one has to be negative or both has to be negative.both can not be positive
so,option 1 is the answer
Using factor method 4x² + 8x + 3 = 4x² + 6x + 2x + 3 = 0
2x(2x + 3) + 1(2x + 3) = 0
(2x + 1)(2x + 3) = 0
Then x = – 1/2 or – 3/2
4x^2+8x+3=0 => 4(x+1)^2-1=0 => x+1=1/2 and -1/2 => x= -1.5 and -0.5
Problem Loading...
Note Loading...
Set Loading...
Descartes' Rule of Signs says it can't have positive solutions. There was only one choice with all numbers negative.