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quadratic equations ( solving for zeros/roots) trickey

miller1991

New member
Jan 23, 2019
2
−2x^2+3x+20


why is this equation so special it is in standard form, and when i solve for zeros
i factor down to

-1( 2x^2-3x-20)

factor
( 2x^2-3x-20)

factor by grouping 2(-20)=-40 what numbers multiple to equal -40 and add to equal -3

-8(5) = -40
-8+5= -3

and then just to find zeros the next step would be
(x-8) (x+5)
x=8 x=-5


BUT
thats not right apparently the answer is

x=4 x = -5/2

the answer i got here

(x-8) (x+5)
x=8 x=-5

is meant to be -1(2x+5)(x-4)

which gives
x=4 x = -5/2


but like 5+-4 =1 dont even equal -3

apparently if i use the huge equation to solve for zeros that works but
why did (2x+5)(x-4) using this work, and how to i recognize when to use it like that
and i dont even know where they pulled the 5 and -4 from


does anyone know why when solving for zeros
y=-2x^2 +3x+20
is a tricky question



p.s
originally the question was solve zeros for
y=3x+20 -2x^2
but putting it in standard form was easy and i dont think that messes with the final answer



thanks !!!
 

joypav

Active member
Mar 21, 2017
151
factor by grouping 2(-20)=-40 what numbers multiple to equal -40 and add to equal -3

-8(5) = -40
-8+5= -3

and then just to find zeros the next step would be
(x-8) (x+5)
x=8 x=-5
You have the right idea!
You are trying to factor
$-2x^2+3x+20$
Factor out the negative (like you did).
$-(2x^2-3x-20)$
Then you determined the values you are working with are 8 and -5. That tells you how to "re-write" your equation so that is set up in a way that you can factor by grouping.
$2x^2-3x-20 = 2x^2+5x-8x-20$
Now factor by grouping.
$2x^2+5x-8x-20 = x(2x+5)-4(2x+5)=(x-4)(2x+5)$
Giving,
$x=4, x=-\frac{5}{2}$