Question

solve the following quadratic equations by factoring: 2x^2-3x=x^2+18

Answer 1

First of all we need to bring it into the standard form of quadratic equation and compare to find a,b and c
The standard form is,
\[ax^2+bx+c=0\]

Then we need to use the quad formula which is,

\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

The other methods we can use here are as follows:

1)Breaking the middle term

2)Completing the whole square.

Answer 2

but my problem is exactly where to start

Answer 3

X(X-3) + 6(X-3) = 0

Answer 4

X=3

Answer 5

Um,The first step is to bring it into standard form,

\[ax^2+bx+c=0\]

Answer 6

so we would have............

Answer 7

2x^2-3x=x^2+18

Subtract both sides by (x^2+18)

Answer 8

Sorry is X(X+3) - 6(X+3) = 0
X= +/- 3

Answer 9

\[x^2-3x-18=0\]

Answer 10

Yeah right :)

Answer 11

so now im stuck again........hmmph

Answer 12

Can you use the quadratic formula ?

Answer 13

we divide by x ???

Answer 14

Think of two numbers when we add or subtract we get -3 and when we multiply we get -18

Answer 15

i know its 6 but what would the work looklike

Answer 16

What do you mean by its 6 ?

Answer 17

I said two numbers.

Answer 18

-3 and 6

Answer 19

Ok so here it is,

\[x^2-6x+3x-18=0\]

Answer 20

Now take common from first two terms and then last two terms.

Answer 21

\[so -6x+3x would be together on 1 term\]

Answer 22

No like,

x(x-6) + 3 (x-6)=0

Answer 23

Either,

x-6=0

Or,

x+3=0

Answer 24

Now solve and get the values of x :)

Answer 25

thank you so much for the help