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4 years ago

2x^2-3x-9 please help

2 answers:

8 0

Im assuming you mean 2x^2-3x-9=0 ?
If so, you have to factorise the equation
(2x-3) (x+3) = 0

2x - 3 = 0
2x = 3
x = 1.5

x + 3 = 0
x = -3

x is either 1.5 or -3

natali 33 [55]4 years ago

8 0

So factor

if you have ax^2+bx+c then to factor find
x+y=b
xy=ac
so

a=2
b-03
c=-9
2 times -9=-18

factor -18
-1,18
-2, 9
-3, 6

now add and find the one that adds to -3

-1+18=17
-2+9=7
-3+ 6=3, could work, if signs are reversed

so the numbers are -6 and 3

split -3 up
2x^2-6x+3x-9
group
(2x^2-6x)+(3x-9)
undistribute
(2x)(x-3)+(3)(x-3)
reverse distributive property
ab+ac=a(b+c)
(2x+3)(x-3)

so we can't do anything else because it isn't an equation so it has non meaning

if it does equal zero then
xy=0 then x and/or y=0 so

(2x+3)(x-3)=0 so
2x+3=0 and x-3=0 so

2x+3=0
subtract 3 from both sides
2x=-3
divide by 2
x=-1.5

x-3=0
add 3 to both sides
x=3

x=-1.5 or 3

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As we have all the answers now.

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So,

Equal to 1:

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Keywords: Fractions, decimals

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